Exercise 6.3 — Triangles

AA, SSS and SAS similarity criteria — choosing the right one, then using it in proofs with medians, altitudes and bisectors

16 questions✓ Free · step-by-step

16 marksNCERT Cl-10 Maths, Ex 6.3, Q1 (Fig. 6.34)
State which pairs of triangles are similar, name the criterion, and write each pair in symbolic form. (i) △ABC with ∠A = 60°, ∠B = 80°, ∠C = 40°; △PQR with ∠P = 60°, ∠Q = 80°, ∠R = 40°. (ii) △ABC with AB = 2, BC = 2.5, CA = 3; △QRP with QR = 4, RP = 5, PQ = 6. (iii) △LMP with LM = 2.7, MP = 2, PL = 3; △DEF with DE = 4, EF = 5, FD = 6. (iv) △MNL with MN = 2.5, ML = 5 and ∠M = 70°; △QPR with QP = 5, QR = 10 and ∠Q = 70°. (v) △ABC with AB = 2.5, BC = 3 and ∠A = 80°; △DEF with DF = 5, EF = 6 and ∠F = 80°. (vi) △DEF with ∠D = 70°, ∠E = 80°; △PQR with ∠Q = 80°, ∠R = 30°.
Hint. Let the data pick the criterion: three angles → AA/AAA; three sides → SSS; two sides and an angle → SAS, but only if that angle is the included one.
(i) Angles only, so test AAA. ∠A = ∠P = 60°, ∠B = ∠Q = 80°, ∠C = ∠R = 40°. All three pairs match in the order given.

✦ △ABC ~ △PQR (AAA similarity)

(ii) Three sides, so test SSS. Match shortest with shortest: AB = 2 with QR = 4, BC = 2.5 with RP = 5, CA = 3 with PQ = 6. 2/4 = 0.5, 2.5/5 = 0.5, 3/6 = 0.5 — all equal.

✦ △ABC ~ △QRP (SSS similarity). Note the order is QRP, not PQR — read it off the ratios.

(iii) Three sides again. Sort each triangle: △LMP has 2, 2.7, 3 and △DEF has 4, 5, 6. 2/4 = 0.5, 2.7/5 = 0.54, 3/6 = 0.5. The middle ratio is out of line, so no correspondence can work.

✦ Not similar

(iv) Two sides and an angle — check whether the angle is included. In △MNL the angle at M lies between MN and ML, and those are exactly the two sides given. In △QPR the angle at Q lies between QP and QR, again the two given sides. So the angle is the included one in both. MN/QP = 2.5/5 = 0.5 and ML/QR = 5/10 = 0.5, and ∠M = ∠Q = 70°.

✦ △MNL ~ △QPR (SAS similarity)

(v) Two sides and an angle — but look where the angle is. The given sides of △ABC are AB and BC, which meet at B; the given angle is at A. So the 80° is not included between them. (In △DEF the 80° at F does sit between DF and EF, but one triangle failing the condition is enough.) With no included angle there is no criterion to apply, and the two triangles need not be similar.

✦ Not similar — the data does not support any criterion

(vi) Angles, but not the same ones. Fill in the missing angles using the 180° sum: in △DEF, ∠F = 180° − 70° − 80° = 30°; in △PQR, ∠P = 180° − 80° − 30° = 70°. Now ∠D = ∠P = 70°, ∠E = ∠Q = 80°, ∠F = ∠R = 30°.

✦ △DEF ~ △PQR (AA similarity)
Where students slip. Answering "similar by SAS" to part (v). It is the one part designed to fail, and it fails on placement, not on arithmetic — the ratios would be fine if the angle were in the right place.
Another way. For (iii) there is a faster disqualifier: the ratio of the largest sides is 3/6 = 0.5 and of the smallest is 2/4 = 0.5, so if the triangles were similar the middle sides would have to satisfy 2.7 = 0.5 × 5 = 2.5. They do not, so you can stop.
23 marksNCERT Cl-10 Maths, Ex 6.3, Q2 (Fig. 6.35)
In the figure △ODC ~ △OBA, with ∠BOC = 125° and ∠CDO = 70°. D and C lie on one line, A and B on a parallel line, and the segments DB and CA cross at O. Find ∠DOC, ∠DCO and ∠OAB.
Hint. ∠DOC and ∠BOC sit on a straight line. Once you have ∠DOC, the angle sum of △ODC gives ∠DCO, and similarity transfers it to △OBA.
Step 1 — Find ∠DOC from the straight line. The points D, O and B lie on one straight line, so ∠DOC and ∠BOC together make a linear pair and must add to 180°. That is why the 125° is useful at all: ∠DOC = 180° − 125° = 55°

Step 2 — Find ∠DCO using the angle sum of △ODC. Two of that triangle's angles are now known, since ∠CDO = 70° is given and ∠DOC = 55° was just found. The third follows: ∠DCO = 180° − 70° − 55° = 55°

Step 3 — Transfer to △OBA using the similarity. △ODC ~ △OBA sets the correspondence O↔O, D↔B, C↔A, which means ∠DCO corresponds to ∠BAO — the angle the question calls ∠OAB. Corresponding angles of similar triangles are equal, and that gives ∠OAB = ∠DCO = 55°

Check. In △OBA the angles are ∠BOA = ∠DOC = 55° (vertically opposite), ∠OBA = ∠ODC = 70°, and ∠OAB = 55°. They sum to 180° ✓

✦ Answer: ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°
Where students slip. Reading ∠OAB off the wrong vertex — matching it with ∠ODC instead of ∠DCO. The similarity statement fixes the pairing, so write out O↔O, D↔B, C↔A before naming any angle.
Another way. Skip the similarity for the last part: DC ∥ AB, so ∠DCO and ∠OAB are alternate angles across the transversal CA and are equal for that reason alone.
33 marksNCERT Cl-10 Maths, Ex 6.3, Q3
Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect at O. Using a similarity criterion, show that OA/OC = OB/OD.
Hint. The parallel sides give you alternate angles at both ends of each diagonal, and the diagonals give vertically opposite angles at O.
Step 1 — Pick the two triangles the diagonals create. Where AC and BD cross they cut the trapezium into four triangles; the two we want are △AOB and △COD, because those are the ones whose sides are the four segments named in the result.

Step 2 — Collect equal angles. AB ∥ DC with AC as a transversal gives ∠OAB = ∠OCD (alternate angles). AB ∥ DC with BD as a transversal gives ∠OBA = ∠ODC (alternate angles). Both come from the single parallel pair — that is the only thing the word trapezium gives us to work with.

Step 3 — Apply AA. Two pairs of equal angles is enough, so △AOB ~ △COD

Step 4 — Read off the sides in the correspondence A↔C, O↔O, B↔D. OA/OC = OB/OD

✦ Answer: Proved — the alternate angles come from the one parallel pair, and AA does the rest.
Where students slip. Using vertically opposite angles at O as one of the two pairs and then stopping. That is legitimate but you still need a second pair, and the alternate angles are the natural source.
Another way. Exercise 6.2 Q9 asks for this same result before similarity criteria are available, and there it needs a constructed parallel through O. Comparing the two proofs shows how much work the criteria save.
43 marksNCERT Cl-10 Maths, Ex 6.3, Q4 (Fig. 6.36)
In the figure QR/QS = QT/PR and ∠1 = ∠2, where ∠1 is the angle at Q and ∠2 the angle at R in △PQR, and S lies on QR. Show that △PQS ~ △TQR.
Hint. The equal angles first tell you something about the *sides* of △PQR. Use that to rewrite the given ratio so both triangles appear in it.
Step 1 — Convert the equal angles into equal sides. In △PQR, ∠1 = ∠2 means ∠PQR = ∠PRQ. Sides opposite equal angles are equal, and the side opposite ∠Q is PR while the side opposite ∠R is PQ. So

PQ = PR … (1)

Step 2 — Use (1) to rewrite the given ratio. We are given QR/QS = QT/PR. Replacing PR by PQ,

QR/QS = QT/PQ

Step 3 — Rearrange so each triangle owns one side of the equation. Cross-multiplying gives QR × PQ = QS × QT, and dividing both sides by QT × QR,

PQ/QT = QS/QR … (2)

Step 4 — Find the included angle. ∠PQS and ∠TQR are both the angle at Q, since S lies on QR. So they are equal, and in each triangle that angle lies between the two sides appearing in (2).

Step 5 — Apply SAS similarity. Equal included angle plus proportional including sides gives

△PQS ~ △TQR

✦ Answer: Proved by SAS similarity.
Where students slip. Substituting PQ = PR the wrong way round. The side opposite ∠Q is PR — writing "∠Q = ∠R so PQ = QR" names a side that is not opposite either angle.
Another way. Some prefer to write the ratio as QS/QP = QR/QT and note the same included angle at Q; it is the same SAS argument with the two fractions inverted together, which is legitimate as long as both are inverted.
53 marksNCERT Cl-10 Maths, Ex 6.3, Q5
S and T are points on sides PR and QR of △PQR such that ∠P = ∠RTS. Show that △RPQ ~ △RTS.
Hint. One pair of equal angles is handed to you. Look for a second — is there an angle both triangles share?
Step 1 — Identify the shared angle. S lies on PR and T lies on QR, so the angle at R in △RTS is the very same angle as the angle at R in △RPQ:

∠PRQ = ∠TRS (common angle)

Step 2 — Use the given equality. We are told ∠P = ∠RTS, that is

∠RPQ = ∠RTS

Step 3 — Apply AA. Two pairs of corresponding angles are equal, so the triangles are similar. Reading the correspondence from the pairs — R↔R, P↔T, Q↔S —

△RPQ ~ △RTS

✦ Answer: Proved by AA similarity.
Where students slip. Writing the conclusion as △RPQ ~ △RST. The angle at P was matched with the angle at T, so P must sit where T sits in the statement, which forces RTS.
Another way. You can also finish by computing the third pair: ∠RQP = 180° − ∠R − ∠P and ∠RST = 180° − ∠R − ∠RTS are equal, giving AAA. It is more work for the same conclusion.
63 marksNCERT Cl-10 Maths, Ex 6.3, Q6 (Fig. 6.37)
In the figure, △ABE ≅ △ACD, with D on AB and E on AC. Show that △ADE ~ △ABC.
Hint. Congruence gives you equal sides, not just equal angles. Write down which sides it pairs up, then look for a shared angle.
Step 1 — Extract the equal sides from the congruence. △ABE ≅ △ACD pairs A↔A, B↔C, E↔D. Corresponding sides of congruent triangles are equal, so

AB = AC and AE = AD … (1)

Step 2 — Build a ratio out of them. From (1), AD = AE and AB = AC. Dividing,

AD/AB = AE/AC

Both fractions equal the same number, since their numerators are equal to each other and so are their denominators.

Step 3 — Find the included angle. ∠DAE and ∠BAC are the same angle at A, because D lies on AB and E lies on AC. In △ADE it lies between AD and AE; in △ABC it lies between AB and AC — exactly the sides in the ratio.

Step 4 — Apply SAS similarity. △ADE ~ △ABC

✦ Answer: Proved by SAS similarity.
Where students slip. Treating the congruence as if it directly gave △ADE ≅ △ABC. It does not — △ADE sits inside △ABC and is genuinely smaller; only the *ratio* of sides transfers.
Another way. An angle route works too: the congruence gives ∠ABE = ∠ACD, and AD = AE with AB = AC makes △ADE isosceles in the same way as △ABC, so the base angles match and AA applies.
74 marksNCERT Cl-10 Maths, Ex 6.3, Q7 (Fig. 6.38)
Altitudes AD and CE of △ABC intersect at P (D on BC, E on AB). Show that (i) △AEP ~ △CDP, (ii) △ABD ~ △CBE, (iii) △AEP ~ △ADB, (iv) △PDC ~ △BEC.
Hint. Every part is AA, and every part gets one of its angles free from an altitude. The second angle is either vertically opposite or shared.
The two altitudes give four right angles to draw on: AD ⊥ BC makes ∠ADB = ∠ADC = 90°, and CE ⊥ AB makes ∠AEC = ∠BEC = 90°.

(i) △AEP ~ △CDP ∠AEP = 90° (part of ∠AEC) and ∠CDP = 90° (part of ∠CDA), so they are equal. ∠APE = ∠CPD, because they are vertically opposite at P. Two pairs of angles ⟹ △AEP ~ △CDP by AA.

(ii) △ABD ~ △CBE ∠ADB = ∠CEB = 90°. ∠ABD = ∠CBE, since both are the angle at B — D lies on BC and E lies on AB, so it is the same angle for both triangles. ⟹ △ABD ~ △CBE by AA.

(iii) △AEP ~ △ADB ∠AEP = ∠ADB = 90°. ∠EAP = ∠DAB, because both are the angle at A: P lies on AD, so ray AP is ray AD, and E lies on AB. ⟹ △AEP ~ △ADB by AA.

(iv) △PDC ~ △BEC ∠PDC = ∠BEC = 90°. ∠DCP = ∠ECB, both being the angle at C — P lies on CE, so ray CP is ray CE, and D lies on CB. ⟹ △PDC ~ △BEC by AA.

✦ Answer: All four pairs are similar by the AA criterion.
Where students slip. Claiming ∠APE = ∠CPD are "corresponding angles". They are vertically opposite — corresponding angles need a pair of parallel lines, and there are none here.
Another way. Parts (i) and (iii) share the triangle AEP, so once you have both you get △CDP ~ △ADB free by transitivity — worth noticing if a later question needs it.
83 marksNCERT Cl-10 Maths, Ex 6.3, Q8
E is a point on side AD produced of a parallelogram ABCD, and BE intersects CD at F. Show that △ABE ~ △CFB.
Hint. A parallelogram gives you equal opposite angles and two pairs of parallel sides. The transversal BE across one of those pairs supplies the second angle.
Step 1 — Use the opposite angles of the parallelogram. In parallelogram ABCD the angles at A and C are opposite, so

∠BAE = ∠FCB … (1)

(∠BAE is the angle at A, since E lies on AD produced and ray AE is ray AD.)

Step 2 — Use a pair of parallel sides with BE as transversal. AD ∥ BC in a parallelogram, and E lies on AD produced, so AE ∥ BC. The line BE cuts both, and

∠AEB = ∠CBF (alternate angles) … (2)

(∠CBF is the angle at B in △CFB, and F lies on BE, so ray BF is ray BE.)

Step 3 — Apply AA. From (1) and (2) two pairs of corresponding angles are equal, so

△ABE ~ △CFB

✦ Answer: Proved by AA similarity.
Where students slip. Forgetting that E is on AD *produced* — outside the parallelogram. The angle at A in △ABE is still ∠BAD because ray AE and ray AD are the same ray; students who redraw E inside get a different, wrong figure.
Another way. Instead of the opposite angles, use the other parallel pair: AB ∥ DC gives ∠ABE = ∠BFC as alternate angles, which with ∠AEB = ∠CBF is again AA.
93 marksNCERT Cl-10 Maths, Ex 6.3, Q9 (Fig. 6.39)
△ABC and △AMP are right triangles, right-angled at B and M respectively, sharing the vertex A (M lies on AC, and B lies on AP). Prove that (i) △ABC ~ △AMP, (ii) CA/PA = BC/MP.
Hint. Part (i) is AA with the right angles and the shared angle at A. Part (ii) is just reading the side ratios off the correspondence from part (i).
(i) The similarity.

Step 1 — Use the two right angles. ∠ABC = 90° and ∠AMP = 90°, so ∠ABC = ∠AMP.

Step 2 — Use the shared vertex. ∠BAC and ∠MAP are the same angle at A, because M lies on AC and B lies on AP.

Step 3 — Apply AA. △ABC ~ △AMP

(ii) The ratio.

Step 4 — Write out the correspondence from part (i): A↔A, B↔M, C↔P. Corresponding sides are therefore in the same ratio:

AB/AM = BC/MP = CA/PA

Step 5 — Take the two ends of that chain.

CA/PA = BC/MP

✦ Answer: (i) △ABC ~ △AMP by AA. (ii) CA/PA = BC/MP, read from that similarity.
Where students slip. Pairing CA with MP and BC with PA. In the correspondence A↔A, B↔M, C↔P, the side CA (from C to A) must pair with PA (from P to A), and BC with MP — the letters have to travel in the same order.
Another way. Part (ii) can also be got from the hypotenuse–leg pairing directly: CA and PA are the hypotenuses, BC and MP the sides opposite the shared angle A, and in similar right triangles those two ratios must agree.
104 marksNCERT Cl-10 Maths, Ex 6.3, Q10
CD and GH are the bisectors of ∠ACB and ∠EGF respectively, with D on AB and H on FE, and △ABC ~ △FEG. Show that (i) CD/GH = AC/FG, (ii) △DCB ~ △HGE, (iii) △DCA ~ △HGF.
Hint. Write down everything the given similarity provides, then halve the angles at C and G. Each part is then AA on a smaller pair of triangles.
Step 1 — Unpack △ABC ~ △FEG. The correspondence is A↔F, B↔E, C↔G, so

∠A = ∠F, ∠B = ∠E, ∠ACB = ∠EGF … (1) AB/FE = BC/EG = CA/GF … (2)

Step 2 — Halve the angles at C and G. CD bisects ∠ACB and GH bisects ∠EGF. Since those two angles are equal by (1), their halves are equal too:

∠ACD = ∠DCB = ½∠ACB ∠FGH = ∠HGE = ½∠EGF so ∠ACD = ∠FGH and ∠DCB = ∠HGE … (3)

(iii) △DCA ~ △HGF — take this one first, because part (i) depends on it. ∠A = ∠F from (1), and ∠ACD = ∠FGH from (3). Two pairs of angles ⟹ △DCA ~ △HGF by AA.

(i) CD/GH = AC/FG From the similarity just proved, with correspondence D↔H, C↔G, A↔F:

DC/HG = CA/GF

which is exactly CD/GH = AC/FG.

(ii) △DCB ~ △HGE ∠B = ∠E from (1), and ∠DCB = ∠HGE from (3). Two pairs of angles ⟹ △DCB ~ △HGE by AA.

✦ Answer: All three follow — (iii) and (ii) by AA, and (i) by reading the sides off (iii).
Where students slip. Assuming the bisectors CD and GH are themselves corresponding sides of the original similarity and quoting CD/GH = AB/FE without proof. The ratio for a cevian has to be earned, which is what part (iii) does.
Another way. Since the ratio in (i) equals the common scale factor of the two triangles, it also equals AB/FE and BC/EG. In other words, bisectors of corresponding angles scale by the same factor as the sides — a useful fact to carry forward.
113 marksNCERT Cl-10 Maths, Ex 6.3, Q11 (Fig. 6.40)
In the figure, 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.
Hint. The two perpendiculars give you a pair of right angles. For the second pair, use the isosceles triangle — and remember that E on CB produced means ray CE is ray CB.
Step 1 — Take the right angles from the perpendiculars. AD ⊥ BC gives ∠ADB = 90°, and EF ⊥ AC gives ∠EFC = 90°. So

∠ADB = ∠EFC … (1)

Step 2 — Take the second pair from the isosceles triangle. AB = AC, so the angles opposite them are equal:

∠ABC = ∠ACB

Now read those angles inside the two triangles we care about. In △ABD the angle at B is ∠ABD, which is ∠ABC. In △ECF the angle at C is ∠ECF — and since E lies on CB produced, the ray CE is the same ray as CB, so ∠ECF is the angle between ray CB and ray CA, that is ∠BCA. Hence

∠ABD = ∠ECF … (2)

Step 3 — Apply AA. From (1) and (2),

△ABD ~ △ECF

✦ Answer: Proved by AA similarity.
Where students slip. Treating ∠ECF as an exterior angle and subtracting it from 180°. E is on CB *produced beyond B*, so C still sees the same ray towards B — ∠ECF simply is ∠BCA, with nothing to subtract.
Another way. If you find the ray argument slippery, mark ∠ABC = ∠ACB = θ on the figure first and then label the two triangles' angles; the equality becomes visible before any writing.
125 marksNCERT Cl-10 Maths, Ex 6.3, Q12 (Fig. 6.41)
Sides AB and BC and median AD of △ABC are respectively proportional to sides PQ and QR and median PM of △PQR. Show that △ABC ~ △PQR.
Hint. A median cuts its side in half. Halving BC and QR turns the given three-way proportion into an SSS statement about the two smaller triangles ABD and PQM.
Step 1 — Write down what is given.

AB/PQ = BC/QR = AD/PM … (1)

Step 2 — Use the definition of a median to bring in BD and QM. AD is the median to BC, so D is the mid-point of BC and BD = ½BC. Likewise QM = ½QR. Therefore

BD/QM = (½BC)/(½QR) = BC/QR

The halves cancel, which is why the median's foot inherits the same ratio as the whole side.

Step 3 — Assemble an SSS statement for the smaller triangles. Combining with (1),

AB/PQ = BD/QM = AD/PM

These are the three sides of △ABD and △PQM, so

△ABD ~ △PQM (SSS similarity)

Step 4 — Harvest the angle. Corresponding angles of similar triangles are equal, and B corresponds to Q:

∠ABD = ∠PQM, that is ∠ABC = ∠PQR

(D lies on BC and M lies on QR, so these are the full angles at B and Q.)

Step 5 — Finish with SAS on the original triangles. We now have AB/PQ = BC/QR from (1), with the included angles ∠B and ∠Q equal. So

△ABC ~ △PQR

✦ Answer: Proved — SSS on the half-triangles delivers the angle, and SAS on the full triangles finishes it.
Where students slip. Trying to apply SSS to △ABC and △PQR directly. You are given only two sides and a median, not the third side — AC and PR never appear in the data, so SSS is unavailable at the top level.
Another way. Question 14 asks the same thing with AB, AC and the median instead of AB, BC and the median. The trick there does not work, because the median's foot no longer lies on a side you have been given — which is why that proof needs a construction.
133 marksNCERT Cl-10 Maths, Ex 6.3, Q13
D is a point on side BC of △ABC such that ∠ADC = ∠BAC. Show that CA² = CB · CD.
Hint. A square on one side and a product of two others is the signature of a similarity where one side appears twice. Look for two triangles sharing the angle at C.
Step 1 — Choose the triangles. Compare △BAC (the whole triangle, read from B) with △ADC (the smaller one cut off by AD).

Step 2 — Find the equal angles. ∠BAC = ∠ADC is given. ∠BCA = ∠ACD, because D lies on BC, so the angle at C is common to both triangles.

Step 3 — Apply AA. △BAC ~ △ADC

Step 4 — Read off the sides in the correspondence B↔A, A↔D, C↔C.

BA/AD = AC/DC = BC/AC

Step 5 — Take the pair that contains AC twice. From AC/DC = BC/AC, cross-multiplying gives

AC × AC = BC × DC CA² = CB · CD

✦ Answer: Proved — CA is the geometric mean of CB and CD.
Where students slip. Writing the similarity as △ABC ~ △ADC. That would pair B with D and force AB/AD = BC/DC, which is not what the angles give. The correct pairing has ∠BAC matched to ∠ADC, so A must sit where D sits.
Another way. Once you recognise the pattern, the answer is one line: CA is the mean proportional between CB and CD, so CA/CD = CB/CA. The full proof is still expected in the exam, but the pattern tells you which ratio to aim for.
145 marksNCERT Cl-10 Maths, Ex 6.3, Q14
Sides AB and AC and median AD of △ABC are respectively proportional to sides PQ and PR and median PM of △PQR. Show that △ABC ~ △PQR.
Hint. The median's foot does not lie on either given side this time, so the Q12 trick fails. Double each median to create a parallelogram and turn the median into a full side.
Step 1 — Write down the hypothesis.

AB/PQ = AC/PR = AD/PM … (1)

Step 2 — Double the medians. Produce AD to E so that AD = DE, and join CE. Since D is the mid-point of BC and also the mid-point of AE, the quadrilateral ABEC has diagonals bisecting each other, so it is a parallelogram. Therefore

CE = AB and AE = 2AD

Do the same in the other triangle: produce PM to N with PM = MN and join RN, giving

RN = PQ and PN = 2PM

Step 3 — Rewrite (1) in terms of the new segments. AB/PQ = CE/RN, because CE = AB and RN = PQ. AD/PM = 2AD/2PM = AE/PN. So (1) becomes

CE/RN = AC/PR = AE/PN

Step 4 — Apply SSS to the doubled triangles. Those are the three sides of △AEC and △PNR, so

△AEC ~ △PNR (SSS similarity)

and therefore ∠CAE = ∠RPN.

Step 5 — Repeat on the other half. The same construction with AB and PQ gives △ABE ~ △PQN, so ∠BAE = ∠QPN.

Step 6 — Add the angles. ∠BAC = ∠BAE + ∠CAE = ∠QPN + ∠RPN = ∠QPR

Step 7 — Finish with SAS. We have AB/PQ = AC/PR from (1) with the included angles ∠A and ∠P now known to be equal, so

△ABC ~ △PQR

✦ Answer: Proved — doubling the median converts it into a side, which is what makes SSS available.
Where students slip. Copying the Q12 method and writing BD/QM = BC/QR. Here BC and QR are not given at all — only AB, AC and the median — so there is nothing for that ratio to connect to.
Another way. The construction is worth remembering as a general move: whenever a median blocks a proof, double it. The parallelogram it creates turns the median into a full side of a triangle you can work with.
152 marksNCERT Cl-10 Maths, Ex 6.3, Q15
A vertical pole 6 m long 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.
Hint. The sun's rays arrive at the same angle for both, and both objects stand vertical. That is two equal angles.
Step 1 — Justify the similarity before using it. The pole and the tower are both vertical, so each makes a right angle with the ground. The sun is far enough away that its rays reach both at the same instant along parallel lines, so the angle of elevation is the same at the tip of each shadow. Two equal angles means the two right triangles are similar by AA.

Step 2 — Write the ratio. Corresponding sides are in the same ratio, so height over shadow is the same number for both:

pole height / pole shadow = tower height / tower shadow 6/4 = h/28

Step 3 — Solve. h = 6 × 28 / 4 = 168/4 = 42

Check. The pole is 1.5 times its shadow (6 ÷ 4), and 28 × 1.5 = 42 ✓

✦ Answer: The tower is 42 m tall.
Where students slip. Setting up 6/4 = 28/h, which inverts one side of the proportion and gives 18.67 m. Keep the same quantity on top in both fractions — height over shadow in each, or shadow over height in each.
Another way. Work with the scale factor instead: the tower's shadow is 28/4 = 7 times the pole's, so every corresponding length is 7 times as big, and the tower is 6 × 7 = 42 m.
163 marksNCERT Cl-10 Maths, Ex 6.3, Q16
AD and PM are medians of triangles ABC and PQR respectively, where △ABC ~ △PQR. Prove that AB/PQ = AD/PM.
Hint. This is question 12 in reverse: there the median proportion was given, here it is to be proved. Start from what the similarity gives you and halve.
Step 1 — Unpack the given similarity. △ABC ~ △PQR gives

AB/PQ = BC/QR … (1) and ∠B = ∠Q … (2)

Step 2 — Halve the sides the medians land on. AD is the median to BC, so BD = ½BC; PM is the median to QR, so QM = ½QR. Dividing,

BD/QM = (½BC)/(½QR) = BC/QR

and by (1) this equals AB/PQ. So

AB/PQ = BD/QM … (3)

Step 3 — Apply SAS to the half-triangles. In △ABD and △PQM the sides AB, BD and PQ, QM are in the same ratio by (3), and the angle between them — ∠B in one, ∠Q in the other — is equal by (2). Therefore

△ABD ~ △PQM (SAS similarity)

Step 4 — Read off the third pair of sides. Corresponding sides of similar triangles are in the same ratio, so

AB/PQ = BD/QM = AD/PM

and in particular AB/PQ = AD/PM.

✦ Answer: Proved — medians of similar triangles scale by the same factor as the sides.
Where students slip. Asserting AD/PM = AB/PQ straight from △ABC ~ △PQR. Similarity is stated for *sides*, and a median is not a side, so the result has to be derived rather than quoted.
Another way. The same argument works for any cevian defined by a fixed ratio — angle bisectors (Q10), altitudes, or the segment to a point one-third along a side. Similar triangles scale every such length by the same factor.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (chapter 6 now runs to three exercises — 6.1, 6.2 and 6.3; the proof of Pythagoras' theorem and the theorem on areas of similar triangles, with old exercises 6.4, 6.5 and 6.6, are no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.

All exercises in Triangles

Exercise 6.1 (3 Q)Exercise 6.2 (10 Q)Exercise 6.3 (16 Q)

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