Exercise 6.2 — Triangles

1. 13 marksNCERT Cl-10 Maths, Ex 6.2, Q1 (Fig. 6.17)

   In both parts DE ∥ BC, with D on AB and E on AC. (i) AD = 1.5 cm, DB = 3 cm and AE = 1 cm — find EC. (ii) AE = 1.8 cm, EC = 5.4 cm and DB = 7.2 cm — find AD.

   Hint. DE ∥ BC is the trigger for the Basic Proportionality Theorem. Write AD/DB = AE/EC and put in what you know.

   Since DE ∥ BC in both parts, Theorem 6.1 (the Basic Proportionality Theorem) applies:

   AD/DB = AE/EC

   (i) Find EC.

   Step 1 — Substitute AD = 1.5, DB = 3, AE = 1. 1.5/3 = 1/EC

   Step 2 — The left side is 0.5, so 0.5 × EC = 1 EC = 2

   ✦ EC = 2 cm

   (ii) Find AD.

   Step 1 — Substitute AE = 1.8, EC = 5.4, DB = 7.2. AD/7.2 = 1.8/5.4

   Step 2 — Simplify the right side. 1.8/5.4 = 1/3, because 5.4 is exactly three times 1.8. AD/7.2 = 1/3

   Step 3 — Solve. AD = 7.2 ÷ 3 = 2.4

   ✦ AD = 2.4 cm

   Check. In (i), AD : DB = 1.5 : 3 = 1 : 2 and AE : EC = 1 : 2 ✓. In (ii), AD : DB = 2.4 : 7.2 = 1 : 3 and AE : EC = 1.8 : 5.4 = 1 : 3 ✓.

   Where students slip. Writing AD/DB = AE/AC. The theorem pairs *part with part* (AD/DB = AE/EC) or *whole with whole* (AD/AB = AE/AC). Mixing the two forms is the most common error in this chapter.

   Another way. Use the whole-side form instead. In (i), AB = 1.5 + 3 = 4.5, so AD/AB = 1.5/4.5 = 1/3; then AE/AC = 1/3 gives AC = 3, and EC = 3 − 1 = 2 cm — the same answer by a slightly longer route.

2. 23 marksNCERT Cl-10 Maths, Ex 6.2, Q2

   E and F are points on sides PQ and PR of △PQR. In each case state whether EF ∥ QR. (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm. (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm. (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm.

   Hint. This is the converse of BPT: compute PE/EQ and PF/FR. Equal ratios mean parallel; unequal ratios mean not parallel.

   The test each time is whether E and F divide PQ and PR in the same ratio. If they do, the converse of BPT (Theorem 6.2) forces EF ∥ QR.

   (i) PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4. PE/EQ = 3.9/3 = 1.3 PF/FR = 3.6/2.4 = 1.5 The ratios differ, so EF is not parallel to QR.

   (ii) PE = 4, QE = 4.5, PF = 8, RF = 9. PE/EQ = 4/4.5 = 8/9 PF/FR = 8/9 Equal, so EF ∥ QR.

   (iii) PQ = 1.28, PR = 2.56, PE = 0.18, PF = 0.36.

   Here the whole sides are given, not the second pieces, so subtract first: EQ = PQ − PE = 1.28 − 0.18 = 1.10 FR = PR − PF = 2.56 − 0.36 = 2.20

   Now test: PE/EQ = 0.18/1.10 = 9/55 PF/FR = 0.36/2.20 = 9/55 Equal, so EF ∥ QR.

   ✦ Answer: (i) not parallel (ii) EF ∥ QR (iii) EF ∥ QR

   Where students slip. In (iii), using PE/PQ and PF/PR without subtracting — or worse, mixing forms by comparing PE/EQ with PF/PR. Either compare both in part-to-part form or both in whole-to-whole form.

   Another way. Part (iii) is quicker in whole-to-whole form: PE/PQ = 0.18/1.28 = 9/64 and PF/PR = 0.36/2.56 = 9/64. Equal ratios, same conclusion, and no subtraction needed.

3. 33 marksNCERT Cl-10 Maths, Ex 6.2, Q3 (Fig. 6.18)

   In the figure, LM ∥ CB and LN ∥ CD, where M lies on AB, N on AD and L on AC. Prove that AM/AB = AN/AD.

   Hint. Two parallels, two triangles. Apply BPT once in △ABC and once in △ACD, and notice that both results involve AL/AC.

   Step 1 — Work in △ABC first. M lies on AB, L lies on AC, and LM ∥ CB. By the Basic Proportionality Theorem in its whole-to-whole form,

   AM/AB = AL/AC … (1)

   Step 2 — Now work in △ACD. N lies on AD, L lies on AC, and LN ∥ CD. The same theorem gives

   AN/AD = AL/AC … (2)

   Step 3 — Compare. The right-hand sides of (1) and (2) are the same quantity, AL/AC — the segment AL and the side AC are shared by both triangles. So the left-hand sides must be equal too:

   AM/AB = AN/AD

   ✦ Answer: Proved. Both ratios equal AL/AC, which is why they equal each other.

   Where students slip. Trying to apply BPT to △ABD or to the quadrilateral as a whole. The theorem needs a *triangle* with a line parallel to one of its sides — here that means splitting the figure into △ABC and △ACD, joined along AC.

   Another way. You can run the whole argument in part-to-part form: AM/MB = AL/LC and AN/ND = AL/LC give AM/MB = AN/ND, then add 1 to both sides to convert back to AM/AB = AN/AD.

4. 43 marksNCERT Cl-10 Maths, Ex 6.2, Q4 (Fig. 6.19)

   In the figure, DE ∥ AC and DF ∥ AE, where D lies on AB and E, F lie on BC. Prove that BF/FE = BE/EC.

   Hint. Use DE ∥ AC in △ABC, then use DF ∥ AE in the smaller triangle △ABE. Both give you a ratio equal to BD/DA.

   Step 1 — Apply BPT in △ABC. D lies on AB, E lies on BC, and DE ∥ AC. So

   BD/DA = BE/EC … (1)

   Step 2 — Apply BPT again, this time in △ABE. D lies on AB, F lies on BE, and DF ∥ AE. The same theorem gives

   BD/DA = BF/FE … (2)

   Step 3 — Compare (1) and (2). Both equal BD/DA, since D and A are the same two points in each. Therefore

   BF/FE = BE/EC

   ✦ Answer: Proved — BD/DA acts as the bridge between the two results.

   Where students slip. Applying the second parallel inside △ABC as well. DF ∥ AE lives in △ABE, not △ABC — the parallel side is AE, so the triangle you use must contain AE as a side.

   Another way. Recognising the shape of the argument saves time later: whenever two parallels share a starting vertex like this, both BPT statements produce the same ratio, and equating them is the whole proof.

5. 53 marksNCERT Cl-10 Maths, Ex 6.2, Q5 (Fig. 6.20)

   In the figure, DE ∥ OQ and DF ∥ OR, where D lies on PO, E lies on PQ and F lies on PR. Show that EF ∥ QR.

   Hint. You are asked to *prove* a line parallel, so the converse of BPT will finish the job. First get PE/EQ and PF/FR both in terms of PD/DO.

   Step 1 — Apply BPT in △POQ. D lies on PO, E lies on PQ, and DE ∥ OQ. So

   PD/DO = PE/EQ … (1)

   Step 2 — Apply BPT in △POR. D lies on PO, F lies on PR, and DF ∥ OR. So

   PD/DO = PF/FR … (2)

   Step 3 — Equate. From (1) and (2),

   PE/EQ = PF/FR

   Step 4 — Now reverse the theorem. In △PQR, the line EF cuts PQ at E and PR at F, and it divides those two sides in the same ratio. That is exactly the hypothesis of Theorem 6.2, the converse of BPT, so

   EF ∥ QR

   ✦ Answer: Proved. Both parallels feed the same ratio PD/DO into △PQR, and the converse of BPT converts equal ratios back into a parallel.

   Where students slip. Stopping at PE/EQ = PF/FR and calling it done. Equal ratios are not the conclusion — you must name the converse of BPT to turn them into the parallel the question asks for.

   Another way. The same three steps work with whole-to-whole ratios (PD/PO = PE/PQ and PD/PO = PF/PR), which some students find cleaner because nothing has to be subtracted.

6. 63 marksNCERT Cl-10 Maths, Ex 6.2, Q6 (Fig. 6.21)

   In the figure, A, B and C are points on OP, OQ and OR respectively, with AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

   Hint. Same structure as the previous question — the shared segment is OA/AP this time.

   Step 1 — Use AB ∥ PQ in △OPQ. A lies on OP and B lies on OQ, so BPT gives

   OA/AP = OB/BQ … (1)

   Step 2 — Use AC ∥ PR in △OPR. A lies on OP and C lies on OR, so

   OA/AP = OC/CR … (2)

   Step 3 — Combine. Both right-hand sides equal OA/AP, therefore

   OB/BQ = OC/CR

   Step 4 — Apply the converse. In △OQR the line BC divides OQ and OR in the same ratio, which by Theorem 6.2 means

   BC ∥ QR

   ✦ Answer: Proved — OA/AP is the common quantity, and the converse of BPT delivers the parallel.

   Where students slip. Writing the ratios with the wrong pieces on top, for example OB/OQ = OC/CR. Keep the pattern consistent: near-piece over far-piece in every ratio, or whole over whole in every ratio.

   Another way. Notice this is question 5 with the vertex moved. Recognising a configuration you have already solved is worth as much in the exam as the algebra itself.

7. 73 marksNCERT Cl-10 Maths, Ex 6.2, Q7

   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. (This is the Mid-point Theorem, which you proved differently in Class 9.)

   Hint. Set up △ABC with D the mid-point of AB and DE ∥ BC. You must show AE = EC — so aim to prove AE/EC = 1.

   Step 1 — Set up. Take △ABC with D the mid-point of AB, and let the line through D parallel to BC meet AC at E. We must show that E is the mid-point of AC.

   Step 2 — Use what "mid-point" gives. D is the mid-point of AB, so AD = DB, and therefore

   AD/DB = 1

   Step 3 — Apply Theorem 6.1. DE ∥ BC, so the Basic Proportionality Theorem gives

   AD/DB = AE/EC

   Step 4 — Substitute. The left side is 1, so AE/EC = 1, which means

   AE = EC

   So E is the mid-point of AC — the line has bisected the third side.

   ✦ Answer: Proved. The mid-point makes the first ratio equal to 1, and BPT passes that 1 across to the second ratio.

   Where students slip. Assuming AE = EC somewhere in the working and then "deriving" it. That is circular — the equality is what you are asked to prove, so it may appear only in the final line.

   Another way. The Class 9 proof went the other way, using congruent triangles and a constructed parallelogram. The BPT version is four lines long, which is why the question specifies Theorem 6.1.

8. 83 marksNCERT Cl-10 Maths, Ex 6.2, Q8

   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.

   Hint. This is the converse of the previous question, so the converse of BPT is the tool. Show both ratios equal 1.

   Step 1 — Name the points. In △ABC let D be the mid-point of AB and E the mid-point of AC. Join DE. We must show DE ∥ BC.

   Step 2 — Write down what the two mid-points give. D is the mid-point of AB, so AD = DB and therefore AD/DB = 1. E is the mid-point of AC, so AE = EC and therefore AE/EC = 1.

   Step 3 — The two ratios are equal. Both are 1, so

   AD/DB = AE/EC

   Step 4 — Apply Theorem 6.2. The line DE divides AB and AC in the same ratio. By the converse of the Basic Proportionality Theorem, DE must be parallel to the third side:

   DE ∥ BC

   ✦ Answer: Proved. Two mid-points make both ratios 1, and equal ratios force the parallel.

   Where students slip. Using Theorem 6.1 instead of 6.2. Theorem 6.1 starts from a parallel and produces a ratio; here you start from ratios and want a parallel, so you need the converse.

   Another way. Questions 7 and 8 are a matched pair — one uses the theorem, the other its converse. Reading them together is the clearest way to remember which theorem points in which direction.

9. 93 marksNCERT Cl-10 Maths, Ex 6.2, Q9

   ABCD is a trapezium with AB ∥ DC, and its diagonals AC and BD intersect at O. Show that AO/BO = CO/DO.

   Hint. There is no line parallel to a side of a triangle in the figure yet — so draw one. Through O, draw a line parallel to AB meeting AD.

   Step 1 — Construct. Through O draw a line parallel to AB, meeting AD at E. Since AB ∥ DC is given and OE ∥ AB, all three lines are parallel:

   OE ∥ AB ∥ DC

   Step 2 — Apply BPT in △ADC. E lies on AD, O lies on AC, and EO ∥ DC. So

   AE/ED = AO/OC … (1)

   Step 3 — Apply BPT in △ABD. E lies on AD, O lies on BD, and EO ∥ AB. So

   AE/ED = BO/OD … (2)

   Step 4 — Compare. Both equal AE/ED, therefore

   AO/OC = BO/OD

   Step 5 — Rearrange into the form asked for. Cross-multiplying gives AO × OD = BO × OC, and dividing both sides by BO × OD gives

   AO/BO = CO/DO

   ✦ Answer: Proved — the constructed parallel through O is what makes both triangles usable.

   Where students slip. Trying to finish without the construction. The trapezium on its own contains no triangle with a line parallel to one of its sides, so BPT has nothing to bite on until you draw OE ∥ AB.

   Another way. Once similarity criteria are available (Exercise 6.3 Q3 asks this again), the construction becomes unnecessary: AB ∥ DC gives ∠OAB = ∠OCD and ∠OBA = ∠ODC as alternate angles, so △AOB ~ △COD by AA and the ratio follows at once.

10. 103 marksNCERT Cl-10 Maths, Ex 6.2, Q10

    The diagonals of a quadrilateral ABCD intersect at O, and AO/BO = CO/DO. Show that ABCD is a trapezium.

    Hint. This is question 9 run backwards. Construct the same parallel through O, get a ratio equality, and finish with the converse of BPT.

    Step 1 — Restate the hypothesis in a more usable form. We are given AO/BO = CO/DO. Cross-multiplying and regrouping,

    AO × DO = BO × CO, so AO/OC = BO/OD … (1)

    Step 2 — Construct. Through O draw a line parallel to AB, meeting AD at E, so OE ∥ AB.

    Step 3 — Apply BPT in △ABD. E lies on AD and O lies on BD with EO ∥ AB, so

    AE/ED = BO/OD … (2)

    Step 4 — Bring in the hypothesis. From (1) and (2),

    AE/ED = AO/OC

    Step 5 — Apply the converse of BPT in △ADC. In △ADC the line EO cuts AD at E and AC at O, dividing them in the same ratio. By Theorem 6.2,

    EO ∥ DC

    Step 6 — Conclude. We now have EO ∥ AB (by construction) and EO ∥ DC (just proved). Two lines parallel to the same line are parallel to each other, so

    AB ∥ DC

    A quadrilateral with one pair of parallel sides is a trapezium.

    ✦ Answer: Proved — ABCD is a trapezium.

    Where students slip. Assuming AB ∥ DC at the start in order to prove it. The parallel is the conclusion here, not a given; the only thing you may assume is the ratio equality.

    Another way. Instead of constructing, use similarity: from AO/OC = BO/OD and the vertically opposite angles ∠AOB = ∠COD, the triangles AOB and COD are similar by SAS. That makes ∠OAB = ∠OCD, and equal alternate angles give AB ∥ DC.

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.