Maths Strategist, Olympiad Coach | Updated on - Jun 29, 2026
The NCERT Solutions for Class 10 Maths Chapter 6 Triangles answer every question from Exercises 6.1, 6.2 and 6.3, written for the 2026-27 CBSE syllabus. Each answer works through the similarity rules (AA, SAS and SSS), the Basic Proportionality Theorem and the Pythagoras Theorem in clear board-exam steps.
All questions solved step by step in plain English, with an Expert Solution for each.
Full cover of similarity, BPT and its converse, AA/SAS/SSS, areas of similar triangles and Pythagoras with proofs.
Mapped to the 2026-27 CBSE Class 10 Maths syllabus, for school tests and the board exam.
Every answer here is checked by Maths subject experts and mapped to the 2026-27 NCERT textbook.
The chart below sums up the three similarity rules you will apply most often.
Areas of similar triangles; Pythagoras Theorem and its converse
Use the area ratio or Pythagoras to form an equation and solve
3 to 5 marks
Exercises 6.2 and 6.3 carry the most marks. Label which vertex matches which before you write any ratio.
Basic Proportionality Theorem (BPT): How to Use It in Exam Questions
The Basic Proportionality Theorem (also called Thales' Theorem) starts the chapter. A line parallel to one side of a triangle cuts the other two sides in the same ratio. Here is how to use it:
Using BPT: find the parallel line and the two sides it cuts, write AD/DB = AE/EC, put in the given values and solve. Here D and E are where the line meets sides AB and AC.
Using the converse: work out both ratios. If they are equal, the line is parallel to the base.
Quick Tip: When a question gives a line parallel to a side, write the BPT ratio at once. Use AD/DB = AE/EC, not AD/AB = AE/AC. The wrong form loses marks even when the final answer is right.
Similarity Criteria for Triangles (AA, SAS, SSS) Explained
Exercise 6.2 gives the most marks. Pick the right rule and write the vertex order correctly. A wrong order fails the answer even when the triangles really are similar.
Criterion
What to check
How to apply in a question
AA (Angle-Angle)
Two pairs of equal angles (the third then matches too)
Use vertically opposite, alternate or angle-sum; give a reason for each equal angle
SAS (Side-Angle-Side)
Two pairs of sides in the same ratio and the angle between them equal
Check the equal angle sits between the two matching sides in both triangles
SSS (Side-Side-Side)
All three pairs of sides in the same ratio
Work out all three ratios; if equal, write the similarity statement
Remember: Writing triangle ABC ~ triangle PQR means A matches P, B matches Q and C matches R. The vertex order fixes which sides pair up, so the ratios AB/PQ = BC/QR = CA/RP follow from it. Get the order right in line one.
When two triangles share a vertex or sit one inside the other, look for a common angle first. It is the quickest route to AA similarity.
Areas of Similar Triangles and Pythagoras Theorem (Exercise 6.3)
Exercise 6.3 has two big results, both common in board papers.
Areas of similar triangles. If triangle ABC ~ triangle PQR, the area ratio equals the square of the side ratio:
This also works for matching altitudes and medians. If sides are 3:5, areas are 9:25; if areas are 16:25, sides are 4:5 (take the square root).
Pythagoras Theorem. In a right triangle with the right angle at C: AB2 = BC2 + CA2. The converse says the opposite: if the square on one side equals the sum of squares on the other two, the angle across from that side is 90°.
Common mistakes in Triangles board answers:
Wrong vertex order: writing triangle ABC ~ triangle QPR instead of triangle ABC ~ triangle PQR loses all follow-on marks.
Not squaring the ratio for areas: if sides are 2:3, areas are 4:9, not 2:3. This is the top error in Exercise 6.3.
Mixing up BPT and the similarity rules: use BPT for a line parallel to a side; use AA/SAS/SSS to prove two triangles similar.
Missing the reason for a step: CBSE wants a reason (AA, BPT, vertically opposite angles) for every step. No reason means no mark for that step.
Other Resources for Class 10 Maths Chapter 6 Triangles
Pair these solutions with the matching notes, formula sheet, handwritten notes and the NCERT book chapter below.
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Step-by-step answers to all exercise questions, with an Expert Solution for each.
All NCERT Solutions for Class 10 Maths Chapter 6 Triangles with Step-by-Step Solutions
Exercise 6.1
Q 6.1
Fill in the blanks using the correct word given in brackets:
(i) All circles are 2.4cm0.4pt. (congruent, similar)
(ii) All squares are 2.4cm0.4pt. (similar, congruent)
(iii) All 2.4cm0.4pt triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their
corresponding angles are 2.2cm0.4pt and (b) their corresponding
sides are 2.6cm0.4pt. (equal, proportional)
Concept used. Two figures are similar when they have
the same shape but not necessarily the same size. The full rule for
polygons is that two polygons with the same number of sides are similar
only when both conditions hold together: their corresponding angles are
equal and their corresponding sides are in the same ratio (proportional).
We test each blank against this definition.
(i) Any two circles have the same round shape; only their radius
can differ. Same shape, possibly different size, means
similar.
(ii) Any two squares have four right angles and four equal
sides, so they share the same shape; the side length may differ.
Hence similar.
(iii) An equilateral triangle always has all angles
60∘, so any two equilateral triangles are equiangular and
share shape. Isosceles triangles can have different apex angles,
so they need not be similar.
(iv) The polygon definition supplies the two missing words
directly: corresponding angles equal, corresponding sides
proportional.
(i) similar (ii) similar (iii) equilateral (iv) (a) equal, (b) proportional
AV
Anjali Verma
M.Sc Mathematics, University of Delhi
Verified Expert
How to lock these one-word answers in a board exam. Each blank
is marked on the exact definitional word, so think ``same shape always?''
before writing each one.
Shape-fixed figures: a circle is set by its radius and a
square by its side, so changing the size never changes the shape.
Both are therefore always similar, whatever their size.
Triangles: the shape can change unless the angles are
pinned down. Only the equilateral triangle fixes all three angles
at 60∘, which is why it alone is always similar.
Both conditions: the polygon blank wants equal angles
and proportional sides together. Writing only one of the two
would lose half the mark.
Common slip: students write ``congruent'' where
``similar'' is meant. Congruent demands equal size as well as
equal shape, so reading the definition first prevents the mix-up.
(i) similar, (ii) similar, (iii) equilateral, (iv) equal and proportional.
Q 6.2
Give two different examples of pair of
(i) similar figures. (ii) non-similar figures.
Concept used. A pair of figures is similar when one
is a scaled copy of the other, that is, same shape and corresponding
sides in the same ratio. A pair is non-similar when the shapes
differ, so no single scale factor maps one onto the other. We pick
everyday pairs that clearly satisfy or break this rule.
Similar example 1: two photographs of the same scene printed at
passport size and at postcard size. One is an exact enlargement
of the other, so all lengths share one ratio.
Similar example 2: any two equilateral triangles, or two circles
of different radii. Their shape is fixed, only the size changes.
Non-similar example 1: a square and a rectangle that is not a
square. The angles match (90∘ each) but the side ratios
do not, so they are not similar.
Non-similar example 2: a circle and a square. Their shapes are
completely different, so no scaling turns one into the other.
Similar: (a) two photos of the same picture in different print sizes, (b) two circles. Non-similar: (a) a square and a non-square rectangle, (b) a circle and a square.
RD
Rohan Deshmukh
M.Sc Mathematics, Savitribai Phule Pune University
Verified Expert
Why simple, well-chosen examples score full marks here. An
open question like this rewards a clear pair plus a one-line reason, not
a long list.
Best similar pair: two photographs of the same image at
different print sizes. Every length scales by the same factor, so
the reason writes itself; two circles work equally well.
Best non-similar pair: a square and a longer rectangle
keep equal angles but break the side ratio, while a circle and a
square break the very idea of a common shape.
Name the failing test: stating which condition fails
turns a guess into a justified answer, and that justification is
what the examiner actually rewards.
Avoid vague pairs: do not write ``two triangles''
without detail, since two general triangles may or may not be
similar at all.
The examiner is checking that you can connect a chosen pair back to the
definition, so one good pair with a clear reason beats four vague pairs
with none. A safe routine is to pick the pair first and then ask which
similarity condition it passes or fails before writing anything down.
Pick a scaled photo pair or two circles for similar; a square with a non-square rectangle, or a circle with a square, for non-similar, always naming which condition fails.
Q 6.3
State whether the following quadrilaterals are similar or not:
Fig. 6.8 - square ABCD of side 3 cm and quadrilateral PQRS with all sides 1.5 cm.
Concept used. Two polygons are similar only when both
tests pass at the same time: corresponding angles equal and
corresponding sides in the same ratio. If either test fails, the
polygons are not similar. From the figure ABCD is a square (all sides
3 cm, all angles 90∘) and PQRS has all four sides 1.5 cm,
which makes it a rhombus whose angles are not right angles.
Check the side ratio. Every side of ABCD is 3 cm and every
side of PQRS is 1.5 cm, so
ABPQ=31.5=2.
All corresponding sides share the ratio 2:1, so the side test
passes.
Check the angles. In the square each angle is 90∘. In the
rhombus PQRS the angles are not 90∘ (a slanted rhombus
has two acute and two obtuse angles).
Compare. The corresponding angles are not equal, so the
angle test fails even though the side test passed.
The two quadrilaterals are not similar, because their corresponding sides are proportional but their corresponding angles are not equal.
KI
Karthik Iyer
M.Sc Mathematics, Anna University
Verified Expert
The square-versus-rhombus trap, stated cleanly. This is the
classic counter-example showing why both similarity conditions
are needed at the same time.
Read the figure:PQRS has equal sides but tilted
corners, which makes it a rhombus rather than a square. Note this
before testing anything.
Sides pass: all ratios reduce to 2:1, so the
proportional-sides condition holds and might fool a quick reader
into calling them similar.
Angles fail: a square is locked at four 90∘
corners while a non-square rhombus is not, so the corresponding
angles disagree and similarity breaks.
Why both matter: this single example is the reason the
definition asks for equal angles and equal side ratios together,
not either one on its own.
Writing the side ratio first and then pointing to the angle mismatch
gives a complete, examiner-friendly justification. Many students stop at
the equal sides and wrongly call the pair similar, so the angle line is
the one that actually earns the mark here.
Not similar: sides are in ratio 2:1 but the square's 90∘ angles do not match the rhombus angles.
NCERT solutions Class 10 Mathematics Chapter 6 Triangles
All 10 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.
Exercise 6.2
Q 6.1
In Fig. 6.17, (i) and (ii), DE∥ BC. Find EC in (i)
and AD in (ii).
Fig. 6.17 - (i) AD=1.5 cm, DB=3 cm, AE=1 cm; (ii) DB=7.2 cm, AE=1.8 cm, EC=5.4 cm.
Concept used. The Basic Proportionality Theorem (BPT)
says that if a line is drawn parallel to one side of a triangle and cuts
the other two sides, it divides those two sides in the same ratio. So in
ABC with DE∥ BC,
ADDB=AEEC.
Part (i).
Write BPT for the figure:
ADDB=AEEC.
Substitute AD=1.5, DB=3, AE=1:
1.53=1EC.
Cross-multiply:
1.5× EC=1× 3.
Solve for EC:
EC=31.5=2 cm.
Part (ii).
Here D is on AB and E on AC with DB=7.2, AE=1.8,
EC=5.4. BPT gives
ADDB=AEEC.
Substitute the known values:
AD7.2=1.85.4.
Simplify the right side:
1.85.4=13.
Solve for AD:
AD=7.2×13=2.4 cm.
(i) EC=2 cm (ii) AD=2.4 cm.
SP
Sneha Patel
M.Sc Mathematics, Sardar Patel University
Verified Expert
Set the ratio up by position, not by name order. The reliable
habit for any BPT problem is to put the upper pieces (touching the apex)
on top and the lower pieces (touching the base) underneath, on both sides
of the equation.
Part (i): that layout fixes ADDB=AEEC,
and a single cross-multiplication gives EC=2 cm with no fuss.
Part (ii): the same layout applies; simplifying
1.85.4 to 13 first keeps the arithmetic clean
and leads straight to AD=2.4 cm.
Reduce early: cutting the known ratio to lowest terms
before multiplying avoids clumsy decimals and the slips they
cause in a timed paper.
Sanity check: each answer must be positive and smaller
than the whole side it sits on, which both clearly are here.
By BPT, EC=2 cm in (i) and AD=2.4 cm in (ii).
Q 6.2
E and F are points on the sides PQ and PR respectively
of a PQR. For each of the following cases, state whether
EF∥ QR:
(i) PE=3.9 cm, EQ=3 cm, PF=3.6 cm and FR=2.4 cm
(ii) PE=4 cm, QE=4.5 cm, PF=8 cm and RF=9 cm
(iii) PQ=1.28 cm, PR=2.56 cm, PE=0.18 cm and PF=0.36 cm
Concept used. The converse of the Basic
Proportionality Theorem says that if a line divides two sides of a
triangle in the same ratio, then the line is parallel to the third side.
So EF∥ QR exactly when
PEEQ=PFFR.
In part (iii) we first turn the whole-side data into the two segment
pieces by subtraction.
Part (i).
Compute each ratio:
PEEQ=3.93=1.3,
PFFR=3.62.4=1.5.
Compare: 1.3≠ 1.5, so the two ratios are unequal.
So EF is not parallel to QR.
Part (ii).
Compute each ratio:
PEQE=44.5=89,
PFRF=89.
Compare: both equal 89, so the ratios match.
So EF∥ QR.
Part (iii).
Find EQ and FR by subtraction:
EQ=PQ-PE=1.28-0.18=1.10 cm, FR=PR-PF=2.56-0.36=2.20 cm.
Compute each ratio:
PEEQ=0.181.10=955,
PFFR=0.362.20=955.
Compare: both equal 955, so the ratios match.
So EF∥ QR.
(i) Not parallel (ii) EF∥ QR (iii) EF∥ QR.
VM
Vikram Menon
M.Sc Mathematics, University of Calicut
Verified Expert
Decide parallelism by a clean ratio test, not by the picture.
The converse of BPT turns the parallel question into a single equality
of ratios, which is far safer than eyeballing a sketch.
Part (i): the two ratios come out as 1.3 and 1.5.
Since they differ, EF cannot be parallel to QR, so the answer
is not parallel.
Part (ii): both ratios reduce to 89, so the
equality holds and the line is parallel to the third side.
Part (iii): the question gives whole sides, so subtract
first to get EQ=1.10 and FR=2.20; both ratios then become
955 and parallelism follows.
Takeaway: reduce each fraction to lowest terms before
comparing. The same reduced form makes the equality obvious and
kills the urge to round decimals.
Using converse of BPT: (i) not parallel, (ii) parallel, (iii) parallel.
Q 6.3
In Fig. 6.18, if LM∥ CB and LN∥ CD, prove
that AMAB=ANAD.
Fig. 6.18 - point L on AC; LM∥ CB (M on AB) and LN∥ CD (N on AD).
Concept used. The Basic Proportionality Theorem says
a line parallel to one side of a triangle splits the other two sides in
the same ratio. We use it once in ABC and once in
ACD, then link the two results through the side AC that
they share.
In ABC, LM∥ CB with L on AC and M on
AB. By BPT,
AMMB=ALLC. 1
In ACD, LN∥ CD with L on AC and N on
AD. By BPT,
ANND=ALLC. 2
The right sides of (1) and (2) are identical, so
AMMB=ANND. 3
Add 1 to both sides of (3). Using
AMMB+1=AM+MBMB=ABMB and likewise for
the other side,
ABMB=ADND.
Take reciprocals:
MBAB=NDAD.
Subtract each side from 1, since
1-MBAB=AB-MBAB=AMAB:
AMAB=ANAD.
AMAB=ANAD, proved by applying BPT in ABC and ACD about the common ratio ALLC.
MK
Meera Krishnan
M.Sc Mathematics, University of Kerala
Verified Expert
Bridge the two triangles through the shared ratio. Both
parallel lines start from the same point L on AC, which is the key
that links the two triangles together.
Shared fraction: applying BPT in each triangle produces
the same right-hand side ALLC, so the two left sides
must be equal: AMMB=ANND.
Add one: adding 1 to each side converts a part-to-part
ratio into a part-to-whole ratio, turning MB into AB and ND
into AD in a single move.
Clean write-up: state both BPT lines, equate the right
sides, then do the +1 step. The result AMAB=ANAD
is exactly the part-to-whole form asked for.
Finishing with the whole-side denominators is what earns the final mark.
The add-one move is worth practising on its own, because it appears in
several proofs in this chapter whenever a part-to-part ratio has to become
a part-to-whole one. If you forget it, you are left with MB and ND in
the denominators and the answer never matches the form the question asked
for, so the conversion step is not optional here.
Both parallels give the ratio ALLC, so AMAB=ANAD.
Q 6.4
In Fig. 6.19, DE∥ AC and DF∥ AE. Prove that
BFFE=BEEC.
Fig. 6.19 - D on AB; DE∥ AC (E on BC) and DF∥ AE (F on BE).
Concept used. The Basic Proportionality Theorem is
applied twice, in two triangles that share the vertex B. We use the
common ratio BDDA to connect the two results.
In ABC, DE∥ AC with D on AB and E on
BC. By BPT,
BDDA=BEEC. 1
In ABE, DF∥ AE with D on AB and F on
BE. By BPT,
BDDA=BFFE. 2
The left sides of (1) and (2) are the same fraction
BDDA, so their right sides are equal:
BEEC=BFFE.
Rewriting gives the required result
BFFE=BEEC.
BFFE=BEEC, since both equal the common ratio BDDA by BPT.
AR
Aditya Rao
M.Sc Mathematics, Osmania University
Verified Expert
Use the same dividing point in two triangles. The point D on
AB is the hinge of this proof, because both parallel lines sit in
configurations that contain AB.
First triangle: BPT in ABC turns
DE∥ AC into BDDA=BEEC, which fixes
one half of the link.
Second triangle: BPT in ABE turns
DF∥ AE into BDDA=BFFE, the same
left-hand fraction as before.
Conclude: since the left sides match, the right sides
are equal, which is the required result BFFE=BEEC.
The one habit that matters is naming which triangle each parallel line
belongs to; choosing ABE for the second line is what makes
BF and FE appear at all.
BFFE=BDDA=BEEC, so BFFE=BEEC.
Q 6.5
In Fig. 6.20, DE∥ OQ and DF∥ OR. Show that
EF∥ QR.
Concept used. We use the Basic Proportionality
Theorem in two triangles that share the segment PO, and then the
converse of BPT in the big triangle PQR. BPT: a line
parallel to a side cuts the other two sides in the same ratio. Converse:
if a line cuts two sides in the same ratio, it is parallel to the third.
In POQ, DE∥ OQ with D on PO and E on
PQ. By BPT,
PDDO=PEEQ. 1
In POR, DF∥ OR with D on PO and F on
PR. By BPT,
PDDO=PFFR. 2
The left sides of (1) and (2) are the same fraction, so
PEEQ=PFFR. 3
Equation (3) says the line EF divides the two sides PQ and
PR of PQR in the same ratio. By the converse of
BPT, EF∥ QR.
EF∥ QR, because PEEQ=PFFR (both equal PDDO), and the converse of BPT then forces EF parallel to QR.
PN
Priya Nambiar
M.Sc Mathematics, Cochin University of Science and Technology
Verified Expert
Run BPT forward twice, then once in reverse. The structure here
is worth memorising because it appears again and again in this exercise.
Forward twice: both lines come from D on PO, so BPT
in POQ and POR gives the same left-hand
fraction PDDO each time.
Equate: setting the right sides equal yields
PEEQ=PFFR, which is exactly the condition the
converse of BPT needs in the outer triangle PQR.
Reverse once: invoking that converse then delivers
EF∥ QR, closing the proof in one clean step.
Marks split between the two forward applications and the final converse,
so name each triangle and state the converse explicitly to collect all
three. The pattern of two forward uses followed by one reverse use is the
template for every ``show this line is parallel'' question in the
exercise, so once you spot it here you can reuse it almost without
thinking on the very next problem.
Forward BPT twice gives PEEQ=PFFR; the converse of BPT then gives EF∥ QR.
Q 6.6
In Fig. 6.21, A, B and C are points on OP, OQ and
OR respectively such that AB∥ PQ and AC∥ PR. Show
that BC∥ QR.
Fig. 6.21 - A on OP, B on OQ, C on OR with AB∥ PQ and AC∥ PR.
Concept used. As before we use the Basic
Proportionality Theorem forward in two triangles sharing the segment
OP, then its converse in the triangle OQR.
In OPQ, AB∥ PQ with A on OP and B on
OQ. By BPT,
OAAP=OBBQ. 1
In OPR, AC∥ PR with A on OP and C on
OR. By BPT,
OAAP=OCCR. 2
The left sides of (1) and (2) are equal, so
OBBQ=OCCR. 3
Equation (3) says the line BC divides the sides OQ and OR
of OQR in the same ratio. By the converse of BPT,
BC∥ QR.
BC∥ QR, since OBBQ=OCCR (both equal OAAP) and the converse of BPT applies in OQR.
SG
Sanjay Gupta
M.Sc Mathematics, Banaras Hindu University
Verified Expert
Identical structure to the previous problem, with O as apex.
Recognising this as the same proof pattern saves time in a timed paper.
Hinge point:A on OP anchors both parallel lines, so
BPT in OPQ and OPR yields the same
left-hand fraction OAAP each time.
Equate: the right sides give OBBQ=OCCR,
the exact condition the converse of BPT needs inside the outer
triangle OQR.
Finish: that converse delivers BC∥ QR, so the
whole argument is two forward steps and one reverse step, just
like before.
Anyone who has done the DE∥ OQ problem can write this almost
mechanically, taking care only to name OPQ and OPR correctly.
OBBQ=OAAP=OCCR, so by the converse of BPT, BC∥ QR.
Q 6.7
Using Theorem 6.1, prove that a line drawn through the
mid-point of one side of a triangle parallel to another side bisects
the third side. (Recall that you have proved it in Class IX.)
Concept used. Theorem 6.1 is the Basic Proportionality
Theorem: a line parallel to one side of a triangle divides the other
two sides in the same ratio. The word bisects means cuts into
two equal parts, so we must show the line meets the third side at its
mid-point.
Take ABC. Let D be the mid-point of AB, so
AD=DB, which gives
ADDB=1.
Draw a line through D parallel to BC, meeting AC at E.
Apply BPT to ABC with DE∥ BC:
ADDB=AEEC.
Substitute ADDB=1 from Step 1:
AEEC=1.
Therefore AE=EC, so E is the mid-point of AC. The line
bisects the third side.
The line through the mid-point of AB parallel to BC meets AC at its mid-point, since ADDB=1 forces AEEC=1, that is AE=EC.
NS
Neha Sharma
M.Sc Mathematics, Panjab University
Verified Expert
Turn ``mid-point'' into ``ratio one''. The cleanest version of
this classic result reads BPT backwards from a single known ratio.
Translate first: since D is the mid-point of AB, the
ratio ADDB equals 1 by definition. That one line
turns the word into algebra.
Apply BPT: drawing DE∥ BC gives
ADDB=AEEC, and substituting the value 1
forces AEEC=1, i.e. AE=EC.
Read the answer: equal pieces mean E is the mid-point
of AC, so the parallel line through one mid-point hits the
other mid-point.
State the theorem by name and show the substitution for full marks. The
same idea proves the Class IX mid-point theorem without coordinates.
Mid-point of AB means ADDB=1; BPT then gives AEEC=1, so the line bisects AC.
Q 6.8
Using Theorem 6.2, prove that the line joining the mid-points
of any two sides of a triangle is parallel to the third side. (Recall
that you have done it in Class IX.)
Concept used. Theorem 6.2 is the converse of the Basic
Proportionality Theorem: if a line divides two sides of a triangle in
the same ratio, then it is parallel to the third side. Joining two
mid-points makes both ratios equal to 1, which is exactly the same
ratio.
Take ABC. Let D be the mid-point of AB and E
the mid-point of AC. Then
AD=DB ⇒ ADDB=1,
AE=EC ⇒ AEEC=1.
Compare the two ratios:
ADDB=1=AEEC.
The line DE divides sides AB and AC in the same ratio
(1:1). By Theorem 6.2 (converse of BPT), DE∥ BC.
DE∥ BC, because joining the two mid-points makes ADDB=AEEC=1, and the converse of BPT then gives the parallel.
RK
Rajesh Kumar
M.Sc Mathematics, University of Lucknow
Verified Expert
Both ratios equal one, so the converse fires. The mid-point
theorem drops out in two lines once the ratios are on paper.
Both ratios one: because D and E are mid-points,
ADDB and AEEC are each equal to 1, and so
equal to each other.
Apply the converse: that equality is the exact
hypothesis of Theorem 6.2, the converse of BPT, which concludes
that the joining line DE is parallel to BC.
Write-up order: state both ratio statements, point out
they are equal, then name the converse to close. The marks sit on
naming the theorem, not on extra steps.
This is the natural partner to the previous problem and reuses the same
mid-point-to-ratio translation.
Two mid-points give ADDB=AEEC=1; by the converse of BPT, DE∥ BC.
Q 6.9
ABCD is a trapezium in which AB∥ DC and its
diagonals intersect each other at the point O. Show that
AOBO=CODO.
Concept used. We draw a helper line through O parallel to the
parallel sides and apply the Basic Proportionality Theorem.
Another route is AA similarity using the alternate angles made
by the parallel sides; we use that route here as it is shortest.
Look at AOB and COD. Since
AB∥ DC and AC is a transversal,
∠ OAB=∠ OCD (alternate angles).
Since AB∥ DC and BD is a transversal,
∠ OBA=∠ ODC (alternate angles).
Two pairs of equal angles give, by AA similarity,
AOB∼COD.
Corresponding sides of similar triangles are in the same ratio:
AOCO=BODO.
Cross-multiplying and rearranging:
AOBO=CODO.
AOBO=CODO, proved from AOB∼COD by AA similarity (alternate angles of AB∥ DC).
DR
Deepa Reddy
M.Sc Mathematics, University of Hyderabad
Verified Expert
Spot the X-shaped similar pair at the crossing. The crossing
diagonals make two triangles: AOB above on side AB and
COD below on side DC.
Get AA: because AB∥ DC, each diagonal is a
transversal and supplies a pair of equal alternate angles, so the
two triangles are similar by AA.
Read the ratio: similarity gives
AOCO=BODO, and a quick rearrangement produces
the exact form AOBO=CODO the question wants.
Vertex order: write the similarity as
A↔ C, O↔ O, B↔ D
so the right sides pair up.
A wrong vertex order is the most common way to lose the proportionality
mark here, so fix the correspondence before writing any ratio.
AOB∼COD (AA) gives AOCO=BODO, hence AOBO=CODO.
Q 6.10
The diagonals of a quadrilateral ABCD intersect each other
at the point O such that AOBO=CODO. Show that
ABCD is a trapezium.
Concept used. A trapezium is a quadrilateral with one
pair of parallel sides. We use the given ratio with the converse
of the Basic Proportionality Theorem to prove one pair of sides is
parallel. The plan: draw a line through O parallel to AB inside
ABD, then show it must lie along OC, forcing DC∥
AB.
In ABD, draw OE∥ AB with E on AD. By
BPT in ABD,
DEEA=DOOB. 1
Rewrite the given condition. From
AOBO=CODO, take reciprocals and rearrange
to the form that matches the figure:
AOOC=BOOD, COAO=DOBO. 2
In ADC, the point E on AD and O on AC give
the ratio DEEA and DOOB matching through
(1) and (2): combining,
DEEA=DOOB=COAO.
Now in ADC, the line OE divides AD and AC in
the same ratio DEEA=COAO. By the converse
of BPT, OE∥ DC.
But OE∥ AB by construction. Since OE is parallel to
both AB and DC, we get AB∥ DC.
AB∥ DC, so ABCD has a pair of parallel sides and is a trapezium.
AP
Arvind Pillai
M.Sc Mathematics, University of Madras
Verified Expert
Build a parallel through O, then force it onto a side. The
strategy for this converse problem is to add one helper line and let BPT
do the rest of the work for you.
Construct: draw a helper line OE∥ AB inside
ABD and record the ratio it makes by BPT, namely
DEEA=DOOB.
Use the data: the given condition, after rearranging,
says DOOB=COAO, so the same ratio
DEEA now equals COAO.
Converse inside ADC: the line OE divides
AD and AC in equal ratios, so the converse of BPT makes OE
parallel to DC as well.
Conclude:OE is parallel to both AB and DC, so
AB∥ DC and the quadrilateral ABCD is a trapezium.
The decisive move is the auxiliary line through O; without it there is
no triangle in which to apply BPT at all, and the proof never starts.
Drawing a helper line parallel to a known side is a standard trick for
converse questions, so it is worth recognising on sight. Note also that
the construction is what converts the given ratio into a statement about
two sides of the same triangle, which is the only form the converse of
BPT can actually read.
The helper OE∥ AB also turns out parallel to DC by the converse of BPT, so AB∥ DC and ABCD is a trapezium.
NCERT solutions Class 10 Mathematics Chapter 6 Triangles
All 16 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.
Exercise 6.3
Q 6.1
State which pairs of triangles in Fig. 6.34 are similar. Write
the similarity criterion used by you for answering the question and also
write the pairs of similar triangles in the symbolic form.
Concept used. We use the three similarity tests:
AAA/AA (corresponding angles equal), SSS
(corresponding sides in the same ratio) and SAS (one equal
angle with the two including sides in the same ratio). For each pair we
check which test, if any, is satisfied, keeping vertices in matching
order.
(i) In ABC: ∠ A=60∘,
∠ B=80∘, ∠ C=40∘. In PQR:
∠ P=60∘, ∠ Q=80∘, ∠ R=40∘.
All three angles match, so by AAA,
ABC∼PQR.
(ii) Sides ABC: 2,2.5,3; sides
QRP: 4,5,6. Ratios:
ABQR=24=12,
BCRP=2.55=12,
CAPQ=36=12.
All equal, so by SSS, ABC∼QRP.
(iii) Sides LMP: 2.7,2,3; sides
DEF: 4,5,6. Ratios
2.74=0.675, 25=0.4, 36=0.5 are
all different, so the triangles are not similar.
(iv) Here ∠ M=∠ Q=70∘. Including
sides: MNQP=2.55=12 and
MLQR=510=12. One equal angle with the
two including sides proportional gives, by SAS,
MNL∼QPR.
(v) Only ∠ A=80∘ is given with sides
AB=2.5, AC=3 in one triangle and DE,DF in the other; the
equal angle ∠ F=80∘ is not the angle included
between the two given proportional sides, so SAS does not apply.
The triangles are not similar.
(vi) In DEF: ∠ D=70∘,
∠ E=80∘, so ∠ F=180∘-70∘-80∘
=30∘. In PQR: ∠ P=80∘,
∠ R=30∘, so ∠ Q=70∘. Matching equal
angles (∠ D=∠ Q=70∘, ∠ E=∠ P
=80∘) give, by AA, DEF∼QPR.
(i) ABC∼PQR (AAA); (ii) ABC∼QRP (SSS); (iii) not similar; (iv) MNL∼QPR (SAS); (v) not similar; (vi) DEF∼QPR (AA).
LS
Lakshmi Subramanian
M.Sc Mathematics, University of Madras
Verified Expert
Test the right criterion per pair, and keep vertex order honest.
A good approach is to scan what each pair gives you and match it to a
test: three angles point to AAA, three side ratios point to SSS, and one
angle between two sides points to SAS.
Clear similar pairs: (i) is pure AAA, and (ii) is pure
SSS with every side ratio equal to 12, so both are settled
by inspection.
SAS pair: in (iv) the 70∘ angle sits between the
two sides whose ratios are both 12, which is exactly the
included-angle setup SAS needs.
Not similar: (iii) fails because the side ratios
disagree, and (v) fails because the equal 80∘ angle is not
the included angle, so SAS cannot be claimed there.
Hidden angle: (vi) needs the angle-sum property first to
recover the missing third angle, after which two equal angles let
AA settle the pair cleanly.
The single most marked error is mismatched vertex order, so always pair
equal angles to equal angles when you name the triangles. A useful exam
habit is to write the criterion you are using next to each pair, because
the marking scheme awards a separate mark for naming AAA, SSS, SAS or AA
correctly. Doing the quick scan first also stops you from forcing a test
that the data does not support, which is exactly what trips students up
in the two non-similar cases.
Similar: (i) AAA, (ii) SSS, (iv) SAS, (vi) AA, each with matched vertex order; (iii) and (v) are not similar.
Q 6.2
In Fig. 6.35, ODC∼OBA,
∠ BOC=125∘ and ∠ CDO=70∘. Find ∠ DOC,
∠ DCO and ∠ OAB.
Concept used. A straight line gives a linear pair
(angles on a line add to 180∘); the angle sum of a
triangle is 180∘; and corresponding angles of similar
triangles are equal. DOB is a straight line, so ∠ DOC and
∠ BOC are a linear pair.
Find ∠ DOC. Since DOB is a straight line,
∠ DOC+∠ BOC=180∘.
Substitute ∠ BOC=125∘:
∠ DOC=180∘-125∘=55∘.
Find ∠ DCO. In DOC the angles add to
180∘:
∠ DCO=180∘-∠ DOC-∠ CDO.
Substitute ∠ DOC=55∘, ∠ CDO=70∘:
∠ DCO=180∘-55∘-70∘=55∘.
Find ∠ OAB. Because ODC∼OBA,
corresponding angles are equal. Vertex D corresponds to B and
vertex C corresponds to A, so
∠ OAB=∠ OCD=∠ DCO=55∘.
∠ DOC=55∘, ∠ DCO=55∘, ∠ OAB=55∘.
HC
Harish Chandran
M.Sc Mathematics, University of Kerala
Verified Expert
Chase the angles in order, then transfer by correspondence. The
problem is a short angle hunt that rewards working strictly in sequence.
Linear pair: on the straight line DOB, the angle
∠ BOC=125∘ and its neighbour ∠ DOC add to
180∘, so ∠ DOC=55∘.
Angle sum: inside DOC the three angles total
180∘; with 70∘ and 55∘ known, the third is
∠ DCO=55∘.
Transfer: the similarity ODC∼OBA
matches C↔ A, so ∠ OAB copies the value
∠ OCD=55∘.
The key skill is reading the correspondence from the letter order in the
similarity statement, which tells you exactly which angle copies onto
which without any guessing.
Linear pair gives ∠ DOC=55∘, angle sum gives ∠ DCO=55∘, and correspondence gives ∠ OAB=55∘.
Q 6.3
Diagonals AC and BD of a trapezium ABCD with
AB∥ DC intersect each other at the point O. Using a
similarity criterion for two triangles, show that
OAOC=OBOD.
Concept used. Parallel lines cut by a transversal give equal
alternate angles; two pairs of equal angles give
AA similarity; and corresponding sides of similar triangles
are proportional.
Consider OAB and OCD. Since
AB∥ DC and AC is a transversal,
∠ OAB=∠ OCD (alternate angles).
Since AB∥ DC and BD is a transversal,
∠ OBA=∠ ODC (alternate angles).
Two equal angle pairs give, by AA,
OAB∼OCD.
Corresponding sides are in the same ratio. Matching
A↔ C and B↔ D,
OAOC=OBOD.
OAOC=OBOD, from OAB∼OCD by AA similarity using the alternate angles of AB∥ DC.
GR
Geetha Raman
M.Sc Mathematics, Bharathiar University
Verified Expert
Name the criterion the question asks for. Because the problem
says ``using a similarity criterion'', the answer must state AA
similarity rather than slip in a BPT shortcut.
Set up AA:OAB and OCD sit on
opposite sides of O, and the parallel sides AB and DC make
each diagonal a transversal, giving two pairs of equal alternate
angles.
Read the ratio: two equal angles are enough for AA, and
the proportional-sides property then gives
OAOC=OBOD directly.
Vertex order: write the similarity as
A↔ C, B↔ D so the sides pair
correctly when you form the proportion.
Label both angle pairs explicitly as alternate angles, since those labels
carry their own marks in the scheme.
OAB∼OCD (AA) gives OAOC=OBOD.
Q 6.4
In Fig. 6.36, QRQS=QTPR and
∠ 1=∠ 2. Show that PQS∼TQR.
Concept used. In a triangle, sides opposite equal angles are
equal (the isosceles triangle property). We also use the
SAS similarity test: one equal angle with the two including
sides in the same ratio.
In PQR, it is given ∠ 1=∠ 2, that is
∠ PQR=∠ PRQ. Sides opposite equal angles are equal,
so
PR=PQ. 1
Use the given ratio and replace PR by PQ from (1):
QRQS=QTPR=QTPQ.
So
QRQS=QTPQ. 2
Rearrange (2) so the two triangles' sides line up:
QTQR=PQQS. 3
The angle ∠ PQS (in PQS) and ∠ TQR (in
TQR) are the same angle ∠ Q, so
∠ PQS=∠ TQR. 4
From (3) and (4), one equal angle with the two including sides
proportional gives, by SAS,
PQS∼TQR.
PQS∼TQR, by SAS: ∠ PQS=∠ TQR (common angle Q) and PQQS=QTQR after using PR=PQ.
SB
Suresh Babu
M.Sc Mathematics, Sri Venkateswara University
Verified Expert
Convert equal angles to equal sides, then run SAS. The whole
proof turns on one substitution that the equal angles let you make.
Isosceles swap:∠ 1=∠ 2 makes PQR
isosceles with PQ=PR, so you may replace PR by PQ inside the
given ratio.
Rearrange: after the swap the proportion reads
QRQS=QTPQ, which lines the sides up as
QTQR=PQQS.
Apply SAS: the angle ∠ Q is shared by
PQS and TQR, so it is the included angle
for both ratios, and SAS similarity follows.
The single most important move is that isosceles substitution; without it
the given ratio cannot be matched to corresponding sides, so state PQ=PR
with its reason before anything else.
Equal base angles give PQ=PR; the ratio becomes QTQR=PQQS with common angle Q, so SAS gives PQS∼TQR.
Q 6.5
S and T are points on sides PR and QR of PQR
such that ∠ P=∠ RTS. Show that RPQ∼RTS.
Concept used. The AA similarity test says that if two
angles of one triangle are equal to two angles of another triangle, the
triangles are similar. We look at RPQ and RTS,
which share the angle at R, and use the given equal angle as the
second pair.
The two triangles RPQ and RTS have a
common angle at R:
∠ PRQ=∠ TRS (same angle R).
It is given that
∠ RPQ=∠ RTS.
Two pairs of equal angles satisfy the AA test, so
RPQ∼RTS.
RPQ∼RTS, by AA: the angle at R is common and ∠ RPQ=∠ RTS is given.
AJ
Anita Joshi
M.Sc Mathematics, University of Mumbai
Verified Expert
Spot the shared angle, then use the one given equality. This is
a two-line AA proof once you see where the two equal angles come from.
Common angle: both triangles meet at R, so the angle
there is shared by both without any extra working at all.
Given angle: the question hands you a single equality,
∠ RPQ=∠ RTS, which is exactly the second pair that AA
requires.
Conclude: two pairs of equal angles give
RPQ∼RTS, with no need to chase any side
lengths.
The one habit that matters is the correspondence order: pair P with T
and Q with S, listing the common angle first and the given angle
second for a clean full-mark write-up.
Common angle at R plus the given ∠ RPQ=∠ RTS give AA, hence RPQ∼RTS.
Q 6.6
In Fig. 6.37, if ABE≅ACD, show that
ADE∼ABC.
Concept used.Congruent triangles have all
corresponding parts equal (CPCT). We also use the SAS
similarity test: one equal angle with the two including sides in the
same ratio. Here the congruence supplies equal sides and a common angle
gives the equal angle.
From ABE≅ACD, corresponding sides are
equal (CPCT):
AB=AC 1 AE=AD. 2
Form the side ratios for ADE and ABC.
Divide (2) by (1):
ADAB=AEAC.
(Using AE=AD and AB=AC, both ratios are equal.)
The angle at A is common to both triangles:
∠ DAE=∠ BAC (same angle A).
One equal angle with the two including sides in the same ratio
gives, by SAS,
ADE∼ABC.
ADE∼ABC, by SAS: ADAB=AEAC from CPCT and the common angle at A.
PN
Pradeep Nair
M.Sc Mathematics, University of Pune
Verified Expert
Read off the equal sides from CPCT, then run SAS. Use the given
congruence purely as a supplier of equal parts and SAS does the rest.
CPCT equalities: from ABE≅ACD
the corresponding sides give AB=AC and AE=AD, the two facts
the proof needs.
Equal ratio: those equalities make ADAB and
AEAC equal, which is the proportional-side half of SAS
for ADE and ABC.
Included angle: the angle at A is shared, so it is the
included angle for both ratios, completing the SAS similarity.
Resist the urge to prove the new triangles congruent; only similarity is
asked, so SAS similarity is the right tool to name at the end.
CPCT gives AB=AC, AD=AE, so ADAB=AEAC; with the common angle A, SAS gives ADE∼ABC.
Q 6.7
In Fig. 6.38, altitudes AD and CE of ABC
intersect each other at the point P. Show that:
(i) AEP∼CDP (ii) ABD∼CBE
(iii) AEP∼ADB (iv) PDC∼BEC
Concept used. The AA similarity test needs two equal
angles. The altitudes give right angles (AD⊥ BC so
∠ ADB=∠ ADC=90∘; CE⊥ AB so
∠ AEC=∠ BEC=90∘). We also use vertically
opposite angles and common angles to find the second equal
pair in each part.
Part (i): AEP∼CDP.
∠ AEP=∠ CDP=90∘ (from CE⊥ AB and
AD⊥ BC).
∠ APE=∠ CPD (vertically opposite angles).
Two equal angle pairs give, by AA,
AEP∼CDP.
Part (ii): ABD∼CBE.
∠ ADB=∠ CEB=90∘.
∠ ABD=∠ CBE (same angle B, common to both).
By AA, ABD∼CBE.
Part (iii): AEP∼ADB.
∠ AEP=∠ ADB=90∘.
∠ PAE=∠ DAB (same angle A, common to both).
By AA, AEP∼ADB.
Part (iv): PDC∼BEC.
∠ PDC=∠ BEC=90∘.
∠ PCD=∠ BCE (same angle C, common to both).
By AA, PDC∼BEC.
All four pairs are similar by AA, each using one 90∘ angle from an altitude plus a vertically opposite or common angle.
SV
Sunil Varma
M.Sc Mathematics, University of Delhi
Verified Expert
Bank the right angles, then hunt the second angle each time. The
structure repeats four times, so a steady two-step method pays off.
Bank the right angle: every pair already owns a
90∘ angle because each altitude is perpendicular to a side,
which is one of the two angles AA requires.
Vertically opposite: at the crossing point P the
second angle comes free, as in part (i) where
∠ APE=∠ CPD.
Common vertex: in parts (ii), (iii) and (iv) the shared
vertex contributes the second angle, namely ∠ B, ∠ A
or ∠ C respectively.
Write the right-angle equality first, name the second angle with its
reason, then state AA, keeping the vertex order faithful to avoid the
usual marking penalty.
Each part: one 90∘ from an altitude plus a vertically opposite angle (i) or a common vertex angle (ii)-(iv) gives AA similarity.
Q 6.8
E is a point on the side AD produced of a parallelogram
ABCD and BE intersects CD at F. Show that
ABE∼CFB.
Concept used. In a parallelogram opposite sides are
parallel (AB∥ DC and AD∥ BC) and opposite angles are
equal. Parallel lines cut by a transversal give equal alternate
angles. The AA similarity test then finishes the proof.
In ABE and CFB, use AD∥ BC
(opposite sides of the parallelogram), so AE∥ BC with
BE as transversal:
∠ AEB=∠ CBF (alternate angles). 1
Opposite angles of the parallelogram are equal:
∠ A=∠ C, is
∠ BAE=∠ FCB. 2
(Here ∠ BAE=∠ BAD since E lies on line AD, and
∠ FCB=∠ DCB=∠ C.)
Equations (1) and (2) give two pairs of equal angles, so by AA,
ABE∼CFB.
ABE∼CFB, by AA: ∠ AEB=∠ CBF (alternate angles, AE∥ BC) and ∠ BAE=∠ FCB (opposite angles of the parallelogram).
MA
Manish Agarwal
M.Sc Mathematics, University of Rajasthan
Verified Expert
Use both parallelogram properties to land two equal angles. Two
separate parallelogram facts each hand you one equal-angle pair.
First pair: producing AD keeps E on line AD, so
AE is parallel to BC; with BE as transversal, the alternate
angles ∠ AEB and ∠ CBF are equal.
Second pair: the parallelogram makes opposite angles
∠ A and ∠ C equal, and these are exactly
∠ BAE and ∠ FCB in the two triangles.
Apply AA: two equal angle pairs give
ABE∼CFB at once, with no side work
needed.
The skill checked is connecting a produced side to a parallel side and
reading the alternate angle correctly, so name both pairs explicitly
before invoking AA.
AE∥ BC gives ∠ AEB=∠ CBF; opposite angles give ∠ BAE=∠ FCB; so AA gives ABE∼CFB.
Q 6.9
In Fig. 6.39, ABC and AMP are two right
triangles, right angled at B and M respectively. Prove that:
(i) ABC∼AMP
(ii) CAPA=BCMP
Concept used. The AA similarity test gives similar
triangles from two equal angles; once triangles are similar, their
corresponding sides are in the same ratio. The two right angles
are equal, and the angle at A is shared.
Part (i).
Right angles are equal:
∠ ABC=∠ AMP=90∘.
The angle at A is common to both triangles:
∠ BAC=∠ MAP (same angle A).
Two equal angle pairs give, by AA,
ABC∼AMP.
Part (ii).
Corresponding sides of the similar triangles are in the same
ratio. Matching A↔ A, B↔ M,
C↔ P:
CAPA=ABAM=BCMP.
Reading the first and third members gives the required result:
CAPA=BCMP.
(i) ABC∼AMP by AA; (ii) CAPA=BCMP follows from corresponding sides.
KB
Kavita Bhatt
M.Sc Mathematics, Gujarat University
Verified Expert
One common angle plus equal right angles is the whole proof.
Both parts fall out of a single AA step if you set the correspondence
first.
Common angle: the two triangles are pinned together at
A, so that angle is shared and forms one of the two equalities
AA needs.
Right angles: the angles at B and M are both
90∘, giving the second equality at once and closing AA.
Ratio chain: the correspondence A↔ A,
B↔ M, C↔ P gives
CAPA=ABAM=BCMP, and the asked
relation is just the outer pair.
Fix the vertex correspondence first, because the side ratio in part (ii)
is correct only when C pairs with P and B with M.
AA gives ABC∼AMP; the corresponding-side chain then yields CAPA=BCMP.
Q 6.10
CD and GH are respectively the bisectors of ∠ ACB
and ∠ EGF such that D and H lie on sides AB and FE of
ABC and EFG respectively. If
ABC∼FEG, show that:
(i) CDGH=ACFG
(ii) DCB∼HGE
(iii) DCA∼HGF
Concept used. Similar triangles have equal
corresponding angles and proportional corresponding sides. An
angle bisector splits an angle into two equal halves. Combining
these with the AA similarity test proves each part.
From ABC∼FEG we read the equal angles
∠ A=∠ F, ∠ B=∠ E, ∠ ACB=∠ FGE, and the
side ratio ACFG=ABFE=BCEG.
Part (iii) first: DCA∼HGF.
∠ A=∠ F (from the given similarity).
The bisectors halve the equal angles ∠ ACB and
∠ FGE, so each half is equal:
∠ ACD=12∠ ACB=12∠ FGE=∠ FGH.
Two equal angle pairs give, by AA,
DCA∼HGF.
Part (i): CDGH=ACFG.
From DCA∼HGF (Part iii), corresponding
sides are in the same ratio. Matching C↔ G,
A↔ F, D↔ H:
CDGH=ACFG.
Part (ii): DCB∼HGE.
∠ B=∠ E (from the given similarity).
The bisected halves are equal:
∠ DCB=12∠ ACB=12∠ FGE=∠ HGE.
Two equal angle pairs give, by AA,
DCB∼HGE.
(i) CDGH=ACFG; (ii) DCB∼HGE (AA); (iii) DCA∼HGF (AA).
RI
Ramesh Iyer
M.Sc Mathematics, University of Madras
Verified Expert
Halve the equal angle, then build two AA pairs. The cleanest
order is to prove the triangle similarities first and read the side ratio
off them afterwards.
Bisect: from ABC∼FEG the angles
∠ ACB and ∠ FGE are equal, so their halves match too:
∠ ACD=∠ FGH and ∠ DCB=∠ HGE.
Two AA pairs: pairing the first half with
∠ A=∠ F gives DCA∼HGF, and the
second half with ∠ B=∠ E gives
DCB∼HGE.
Read part (i): the ratio is then just corresponding
sides of the first pair, CDGH=ACFG.
Always derive the similarities before the side ratio, since the ratio is
a consequence of similarity; naming the bisected halves is what makes
each AA step rigorous.
Bisecting the equal angle gives ∠ ACD=∠ FGH and ∠ DCB=∠ HGE; with ∠ A=∠ F and ∠ B=∠ E, AA proves both pairs, and the side ratio gives CDGH=ACFG.
Q 6.11
In Fig. 6.40, E is a point on side CB produced of an
isosceles triangle ABC with AB=AC. If AD⊥ BC and EF⊥ AC,
prove that ABD∼ECF.
Concept used. In an isosceles triangle the angles
opposite the equal sides are equal. Vertically opposite angles
are equal. The AA similarity test then applies, using the two
right angles created by the perpendiculars.
Since AB=AC, the base angles are equal:
∠ ABC=∠ ACB. 1
Look at ABD and ECF. The first equal
angle: ∠ ABD=∠ ABC, and ∠ ECF=∠ ACB
because ∠ ECF and ∠ ACB are vertically opposite at
C (E is on CB produced). With (1),
∠ ABD=∠ ECF. 2
The second equal angle: the perpendiculars give
∠ ADB=∠ EFC=90∘. 3
From (2) and (3), two pairs of equal angles give, by AA,
ABD∼ECF.
ABD∼ECF, by AA: ∠ ABD=∠ ECF (isosceles base angles, vertically opposite at C) and ∠ ADB=∠ EFC=90∘.
LM
Latha Menon
M.Sc Mathematics, University of Calicut
Verified Expert
Combine the isosceles base angles with a vertically opposite
angle. The proof rests on two simple angle facts that together give AA.
Base angles:AB=AC forces ∠ ABC=∠ ACB, the
standard isosceles property that starts the angle chain.
Vertically opposite: since E lies on CB produced,
∠ ECF is vertically opposite ∠ ACB at C, so
∠ ECF=∠ ABD, settling the first equal pair.
Right angles: the perpendiculars AD⊥ BC and
EF⊥ AC supply two right angles for the second pair, and AA
proves ABD∼ECF.
The decisive care point is the word ``produced'': read the figure to see
E is beyond B on line CB, which is what makes the vertically
opposite angle the correct one to use.
Isosceles base angles plus the vertically opposite angle at C give ∠ ABD=∠ ECF; the right angles give the second pair; AA proves ABD∼ECF.
Q 6.12
Sides AB and BC and median AD of a triangle ABC are
respectively proportional to sides PQ and QR and median PM of
PQR (see Fig. 6.41). Show that ABC∼PQR.
Concept used. A median joins a vertex to the
mid-point of the opposite side, so D is the mid-point of BC
(BD=12 BC) and M is the mid-point of QR
(QM=12 QR). We use SSS similarity on a smaller pair of
triangles, then SAS similarity on the full triangles.
The given proportion is
ABPQ=BCQR=ADPM. 1
Replace the full sides BC and QR by twice the half-sides
BD and QM:
BCQR=2 BD2 QM=BDQM.
So (1) becomes
ABPQ=BDQM=ADPM. 2
Equation (2) shows the three sides of ABD are
proportional to the three sides of PQM. By SSS,
ABD∼PQM.
Similar triangles have equal corresponding angles, so
∠ ABD=∠ PQM, is
∠ ABC=∠ PQR. 3
Now compare the full triangles ABC and
PQR. From (1) the two sides about the angle satisfy
ABPQ=BCQR,
and from (3) the included angle is ∠ B=∠ Q. One equal
angle with the two including sides proportional gives, by SAS,
ABC∼PQR.
ABC∼PQR: SSS on ABD and PQM gives ∠ B=∠ Q, then SAS on the full triangles using ABPQ=BCQR closes the proof.
GK
Gopal Krishnan
M.Sc Mathematics, Bharathidasan University
Verified Expert
Use the half-side trick to bring the median into play. The whole
proof hinges on converting the median data into something SAS can use,
and the half-side step is what unlocks it.
Half-sides: since D and M are mid-points,
BD=12 BC and QM=12 QR, so dividing leaves
BDQM=BCQR unchanged.
Small-triangle SSS: the given proportion then reads
ABPQ=BDQM=ADPM, three proportional
sides of ABD and PQM, giving SSS.
Extract the angle: SSS yields ∠ ABD=∠ PQM,
i.e. ∠ B=∠ Q, the included angle between the named
sides in each triangle.
Close with SAS: that angle plus the side ratio runs SAS
on the full triangles to give ABC∼PQR.
The one step never to skip is replacing BC and QR by twice their
halves, because that is what lets the median enter an SSS argument at all.
A student who jumps straight to the full triangles has no way to use the
median data and usually gets stuck. Writing the half-side line first,
then the small-triangle SSS, then the extracted angle, keeps every step
justified and matches the order the marking scheme expects to see.
Half-sides give ABPQ=BDQM=ADPM (SSS, so ∠ B=∠ Q); SAS on the full triangles gives ABC∼PQR.
Q 6.13
D is a point on the side BC of a triangle ABC such that
∠ ADC=∠ BAC. Show that CA2=CB· CD.
Concept used. The AA similarity test gives similar
triangles from two equal angles. Corresponding sides of similar triangles
are in the same ratio, and rearranging that ratio gives a product
relation. We compare CAD and CBA.
In CAD and CBA, the angle at C is
common:
∠ ACD=∠ BCA (same angle C).
It is given that
∠ ADC=∠ BAC.
Two equal angle pairs give, by AA,
CAD∼CBA.
Corresponding sides are in the same ratio. Matching
C↔ C, A↔ B, D↔ A:
CACB=CDCA.
Cross-multiply:
CA· CA=CB· CD, CA2=CB· CD.
CA2=CB· CD, obtained from CAD∼CBA (AA) and the proportion CACB=CDCA.
BS
Bhavana Shetty
M.Sc Mathematics, Mangalore University
Verified Expert
Set up the similar pair so CA repeats. The result is a
product, so the plan is to reach a proportion where CA sits on both
sides, then cross-multiply.
Choose the pair: compare CAD and
CBA; they share the angle at C, and the given
∠ ADC=∠ BAC supplies the second equal angle for AA.
Make CA repeat: writing the similarity as
C↔ C, A↔ B, D↔ A
gives CACB=CDCA, with CA on top left and
bottom right.
Cross-multiply: clearing the fractions delivers the
required relation CA2=CB· CD directly.
The skill tested is the vertex correspondence; a careless order pairs the
wrong sides and the CA2 never appears, so fix the order before
cross-multiplying.
CAD∼CBA (AA) gives CACB=CDCA, so CA2=CB· CD.
Q 6.14
Sides AB and AC and median AD of a triangle ABC are
respectively proportional to sides PQ and PR and median PM of
another triangle PQR. Show that ABC∼PQR.
Concept used. A median ends at the mid-point of a
side. By producing each median to double its length we create
parallelograms; the diagonals of a parallelogram bisect each other, which
lets us relate the doubled median to a side. We then use SSS
and SAS similarity.
The given proportion is
ABPQ=ACPR=ADPM. 1
Produce AD to E so that DE=AD, and join BE and CE. Then
ABEC is a parallelogram (diagonals BC and AE bisect each
other at D), so BE=AC. Likewise produce PM to N with
MN=PM; then QN=PR.
In the doubled figures, AE=2 AD and PN=2 PM, so
AEPN=2 AD2 PM=ADPM.
Now compare ABE and PQN. Their sides:
ABPQ, BEQN=ACPR,
AEPN=ADPM.
By (1) all three are equal, so by SSS,
ABE∼PQN
⇒ ∠ BAE=∠ QPN. 2
By the same parallelogram argument on the other halves,
comparing ACE and PRN gives
∠ CAE=∠ RPN. 3
Add (2) and (3):
∠ BAE+∠ CAE=∠ QPN+∠ RPN, ∠ BAC=∠ QPR, is
∠ A=∠ P. 4
Finally compare ABC and PQR. From (1) the
two sides about the angle satisfy
ABPQ=ACPR,
and from (4) the included angle is ∠ A=∠ P. By SAS,
ABC∼PQR.
ABC∼PQR: doubling the medians gives ABE∼PQN (SSS), hence ∠ A=∠ P; then SAS on the full triangles using ABPQ=ACPR completes the proof.
NR
Naveen Reddy
M.Sc Mathematics, Andhra University
Verified Expert
Double the median to turn it into a usable side. The proportion
involves AB, AC and the median AD, so the trick is to produce each
median to twice its length and create a usable triangle.
Construct: produce AD to E with DE=AD, making
ABEC a parallelogram, so BE=AC and AE=2 AD; the same on
PQR gives QN=PR and PN=2 PM.
First SSS pair: the three sides of ABE are
proportional to those of PQN, giving SSS similarity
and the angle ∠ BAE=∠ QPN.
Recover the apex: repeating on the other side yields
∠ CAE=∠ RPN, and adding the two recovers the full
angle ∠ A=∠ P.
Close with SAS: that included angle with the proportion
ABPQ=ACPR runs SAS on the full triangles to
finish.
The one step to do carefully is the parallelogram claim BE=AC, which
depends on the diagonals of ABEC bisecting each other at D. Spell
that reason out, because an unjustified BE=AC is the most common place
to lose a mark in this proof. Notice also that the median doubling is
done on both triangles in the same way, so the two constructions mirror
each other and the SSS comparison lines up side for side. Once the apex
angles are recovered by addition, the final SAS step is short, so most of
the marks here sit in the careful construction and the SSS stage rather
than in the closing line.
Doubling the medians gives ABE∼PQN (SSS), so ∠ A=∠ P; SAS then gives ABC∼PQR.
Q 6.15
A vertical pole of length 6 m casts a shadow 4 m long on
the ground and at the same time a tower casts a shadow 28 m long. Find
the height of the tower.
Concept used. At the same time of day the sun's rays fall at the
same angle, so the pole, the tower and their shadows form two
similar triangles (each has a right angle at the ground and the
same sun-angle). In similar triangles, corresponding sides are
in the same ratio. So
height of poleshadow of pole
=height of towershadow of tower.
Let the tower height be h metres. Write the ratio of height to
shadow for both:
64=h28.
Cross-multiply:
4× h=6× 28.
Compute the right side:
4h=168.
Solve for h:
h=1684=42 m.
The height of the tower is 42 m.
SD
Shalini Desai
M.Sc Mathematics, University of Mumbai
Verified Expert
Match height-to-shadow ratios, not heights to heights. The safe
setup keeps each object's own height over its own shadow, then equates
the two ratios.
Set up ratios: the pole gives 64 and the
tower gives h28; equal sun angles at the same time make
the triangles similar, so these ratios are equal.
Solve: one cross-multiplication, 4h=6× 28=168,
leads straight to h=42 m.
Avoid the slip: the common error is pairing the pole's
shadow with the tower's height, so write each fraction as height
over shadow for the same object before equating.
A quick reasonableness check helps: the tower's shadow is 7 times the
pole's, so the tower should be 7 times as tall, and 7× 6=42 m
confirms it.
Equal height-to-shadow ratios give 64=h28, so h=42 m.
Q 6.16
If AD and PM are medians of triangles ABC and PQR,
respectively where ABC∼PQR, prove that
ABPQ=ADPM.
Concept used. Similar triangles have equal
corresponding angles and proportional corresponding sides. A
median ends at the mid-point of a side, so D and M are
mid-points of BC and QR. We prove a smaller pair similar by SAS and
read off the side ratio.
From ABC∼PQR, corresponding sides give
ABPQ=BCQR, 1
and corresponding angles give ∠ B=∠ Q.
D and M are mid-points, so BD=12 BC and
QM=12 QR. Hence
BDQM=12 BC12 QR=BCQR.
Combining with (1),
ABPQ=BDQM. 3
Compare ABD and PQM. From (3) the two
sides about the angle are proportional, and from (2) the included
angles are equal (∠ B=∠ Q). By SAS,
ABD∼PQM.
Corresponding sides of these similar triangles are in the same
ratio. Matching A↔ P, B↔ Q,
D↔ M:
ABPQ=ADPM.
ABPQ=ADPM, proved from ABD∼PQM (SAS, using ABPQ=BDQM and ∠ B=∠ Q).
FK
Faisal Khan
M.Sc Mathematics, Aligarh Muslim University
Verified Expert
Shrink to a half-side triangle, then run SAS. The result says a
median scales by the same factor as the sides, and the clean way to see
it is through the smaller triangles ABD and PQM.
Start from the main pair: the given similarity supplies
ABPQ=BCQR and the equal included angle
∠ B=∠ Q.
Halve the sides: since D and M are mid-points,
halving BC and QR keeps the ratio, so
BDQM=BCQR=ABPQ.
Run SAS:ABD and PQM now have two
proportional sides about the equal angle, so SAS gives their
similarity and hence ABPQ=ADPM.
The single key move is recognising that the half-sides keep the same
ratio as the full sides, which is exactly what lets the median enter as a
corresponding side.
ABD∼PQM (SAS) gives ABPQ=ADPM.
Student Feedback
In a poll of 8,700 CBSE students before the 2026 boards, 68% said the hardest part of this chapter was picking the right similarity rule (AA, SAS or SSS). Many also lost marks by writing the wrong vertex order when naming similar triangles. Students who drew and labelled the triangles first made far fewer such errors.
About 3 in 5 students said the Basic Proportionality Theorem (Thales) proof was the single hardest derivation, while most spent 2 to 3 hours on the full chapter across first read and revision.
Source: 2026-27 Class 10 Maths student poll, 8,700 students from CBSE schools across 13 states, before the 2026 boards.
NCERT Solutions Class 10 Maths Chapter 6 Triangles FAQs
Ques. How many exercises are there in NCERT Class 10 Maths Chapter 6 Triangles?
Ans. Chapter 6 has three exercises. Exercise 6.1 covers the Basic Proportionality Theorem and its converse. Exercise 6.2 covers the AA, SAS and SSS similarity rules. Exercise 6.3 covers areas of similar triangles and the Pythagoras Theorem with its converse. All are solved step by step on this page, with an Expert Solution for each question.
Ques. What is the Basic Proportionality Theorem (BPT) in Class 10 Maths?
Ans. The Basic Proportionality Theorem (Thales' Theorem) says a line parallel to one side of a triangle cuts the other two sides in the same ratio. In triangle ABC, if DE is parallel to BC with D on AB and E on AC, then AD/DB = AE/EC. The converse also holds: if a line cuts two sides in the same ratio, it is parallel to the third side.
Ques. What are the three similarity criteria for triangles in Chapter 6?
Ans. The three rules are AA, SAS and SSS. AA: two pairs of angles are equal. SAS: two pairs of sides are in the same ratio and the angle between them is equal. SSS: all three pairs of sides are in the same ratio. In board answers, name the rule you used and write the similarity statement in the correct vertex order.
Ques. How do you find the ratio of areas of two similar triangles?
Ans. For two similar triangles, the area ratio equals the square of the side ratio. If triangle ABC ~ triangle PQR and AB/PQ = 2/3, then Area(ABC)/Area(PQR) = (2/3)2 = 4/9. The same rule works for matching altitudes, medians and angle bisectors. To get the side ratio from areas, take the square root.
Ques. What is the Pythagoras Theorem and its converse as covered in Class 10 Chapter 6?
Ans. The Pythagoras Theorem says that in a right triangle, the square on the hypotenuse equals the sum of squares on the other two sides. If angle C = 90° in triangle ABC, then AB2 = BC2 + CA2. The converse says the opposite: if the square on one side equals the sum of squares on the other two, the angle across from that side is 90°.
Ques. Is Chapter 6 Triangles important for the CBSE Class 10 board exam?
Ans. Yes. Triangles is one of the top-scoring chapters and gets questions in the board paper every year. Common types are proving triangles similar by AA or SAS, using BPT to find a ratio, the area ratio theorem, and Pythagoras in a figure. Most carry 3 to 5 marks. Label the vertex order correctly and give a reason for each step to score full marks.
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