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**Question 1. **Î”**ABC and **Î”**DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that (i) **Î”

(ii)

(iii) AP bisects âˆ A as well as âˆ D.

(iv) AP is the perpendicular bisector of BC.

**Solution:** (i) In Î”ABD and Î”ACD, we have

AB = AC [Given]

AD = AD [Common]

BD = CD [Given]

âˆ´ Î”ABD â‰Œ Î”ACD [SSS Criteria]

(ii) In Î”ABP and Î”ACP, we have

AB = AC [Given]

âˆ´ AB = AC

â‡’ âˆ B = âˆ C [âˆµ Angle opposite to equal sides are equal]

AP = AP [Common]

âˆ´ Î”ABP â‰Œ Î”ACP [SAS Criteria]

(iii) Since, Î”ABP â‰Œ Î”ACP

âˆ´ Their corresponding parts are congruent.

â‡’ âˆ BAP = âˆ CAP

âˆ´ AP is the bisector of âˆ A . ...(1)

Again, in Î”BDP and Î”CDP, we have

BD = CD [Given]

âˆ DBP = âˆ CDP [Angles opposite to equal sides]

DP = DP [Common]

â‡’ âˆ BDP â‰Œ âˆ CDP

âˆ´ âˆ BDP = âˆ CDP [c.p.c.t.]

â‡’ DP (or AP) is the bisector of âˆ D . ...(2)

From (1) and (2), AP is the bisector of âˆ A as well as âˆ D.

(iv) âˆµ Î”ABP â‰Œ Î”ACP

âˆ´ Their corresponding parts are equal.

â‡’ âˆ APB = âˆ APC

But âˆ APB + âˆ APC = 180Âº [Linear pair]

âˆ´ âˆ APB = âˆ APC = 90Âº

â‡’ AP âŠ¥ BC

â‡’ AP is the perpendicular bisector of BC

**Question 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects âˆ A Solution: **(i) In Î”ABD and Î”ACD, we have

AB = AC [Given]

âˆ B= âˆ C

[Angles opposite to equal sides]

AD = AD [Common]

âˆ´ Î”ABD â‰Œ Î”ACD

â‡’ Their corresponding parts are equal.

âˆ´ BD = CD

â‡’ D is the mid-point of BC or AD bisects BC.

(ii) Since, Î”ABD â‰Œ Î”ACD,

âˆ´ Their corresponding parts are congruent.

â‡’ âˆ BAD = âˆ CAD

â‡’ AD bisects âˆ A.

**Question 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Î”PQR (see figure). Show that (i) Î”ABM â‰Œ Î”PQN (ii) Î”ABC â‰Œ Î”PQR**

**Solution:** In Î”ABC, AM is a median [Given].

âˆ´ BM = (1/2)BC â€¦(1)

In Î”PQR, PN is a median.

âˆ´ QN = (1/2)QR â€¦(2)

âˆµ BC = QR [Given]

âˆ´ (1/2)BC = (1/2)QR

â‡’ BM = QN

[From (1) and (2)]

(i) In Î”ABM and Î”PQN, we have

âˆµ AB = PQ [Given]

AM = PN [Given]

BM = QN [Proved]

âˆ´ Î”ABM â‰Œ Î”PQN [SSS criteria]

(ii) âˆµ Î”ABM â‰Œ Î”PQN

âˆ´ Their corresponding parts are congruent.

â‡’ âˆ B= âˆ Q

Now, in Î”ABC and Î”PQR, we have

âˆ B= âˆ Q [Proved]

AB = PQ [Given]

BC = QR [Given]

âˆ´ Î”ABC â‰Œ Î”PQR

**Question 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles. Solution: **âˆµ BE âŠ¥ AC [Given]

âˆ´ Î”BEC is a right triangle such that âˆ BEC = 90Âº

Similarly, âˆ CFB = 90Âº

Now, in right Î”BEC and right Î”CFB, we have

BE = CF [Given]

BC = CB [common]

âˆ´ Using RHS criteria,

Î”BEC â‰Œ Î”CFB

âˆ´ Their corresponding parts are equal.

â‡’ âˆ BCE = âˆ CBF or âˆ BCA = âˆ CBA

Now, in Î”ABC, âˆ BCA = âˆ CBA

âˆ´ Their opposite sides are equal.

â‡’ AB = AC

âˆ´ Î”ABC is an isosceles triangle.

**Question 5. ABC is an isosceles triangle with AB = AC. Draw AP âŠ¥ BC to show that âˆ B = âˆ C. Solution:** We have AP âŠ¥ BC [Given]

âˆ´ âˆ APB = 90Âº and APC = 90Âº

In Î”ABP and Î”ACP, we have

âˆ APB = âˆ APC [each = 90Âº]

AB = AC [Given]

AP = AP [common]

âˆ´ Using RHS criteria,

Î”ABP â‰Œ Î”ACP

âˆ´ Their corresponding parts are congruent.

â‡’ âˆ B= âˆ C

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