**Q.1Â Â Â Â ABC and 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 **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(ii) **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(iii) AP bisects as well as **

Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(iv) AP is the perpendicular bisector of BC.**

**Sol.**

**(i)** In ABD and ACD, we have

AB = AC [Given]

BD = DC [Given]

and AD = AD [Common]

Therefore by SSS criterion of congruence, we have

** (ii)** In ABP and ACP , we have

AB = AC [Given]

[Since

Â Â Â Â

and Â Â AP = AP [Common]

Therefore By SAS criterion of congruence , we have Â

**(iii)** Since . Therefore ,

AD bisects AP bisects ... (1)

In BDP and CDP, we have

BD = CD Â Â Â Â Â Â Â Â Â [Given]

BP = PC [Since BP = PC]

and DP = DP [Common]

By SSS criterion of congruence, we have

DP bisects AP bisects ... (2)

Combining (1) and (2), we get

AP bisects as well as

**(iv)** Since AP stands on BC

Therefore 180Âº [Linear Pairs]

But [Proved above]

Therefore

Also BP = PC [Proved above]

AP is perpendicular bisector of BC.

**Q.2Â Â Â Â AD is an altitude of an isosceles triangle ABC in which AB = AC. Show thatÂ Â Â Â Â Â Â Â Â Â Â (i) AD bisects BCÂ Â Â Â Â Â Â Â Â Â Â Â Â Â (ii) AD bisects .**

AD is the altitude drawn from vertex A of an isosceles ABC to the opposite base BC so that AB = AC, 90Âº.

Now, in ADB and ADC, we have

Hyp. AB = Hyp. AC [Given]

AD = AD [Common]

and [Since Each = 90Âº]

Therefore By RHS criterion of congruence, we have

BD = DC and

[Since corresponding parts of congruent triangle are equal]

Hence, AD bisects BC, which proves (i) and AD bisects , which proves (ii).

**Q.3Â Â Â Â Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR (see figure). ****Show that :Â Â Â Â Â Â Â Â Â (i) Â Â Â Â Â Â Â Â Â (ii)
**

Two ABC and PQR in which AB = PQ, BC = QR and AM = PN.

Since AM and PN are median of ABC and PQR respectively.

Now, BC = QR [Given]

(Median divides opposite sides in two equal parts)

BM = QN ... (1)

Now , in ABM and PQN we have

AB = PQ [Given]

BM = QN [From (i)]

and AM = PN [Given]

Therefore By SSS criterion of congruence, we have

, which proves (i)

... (2)

[Since, corresponding parts of congruent triangle are equal]

Now, in ABC and PQR we have

AB = PQ [Given]

[From (2)]

BC = QR [Given]

Therefore By SAS criterion of congruence, we have

, which proves (ii)

Â **Q.4Â Â Â Â Â BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.**

**Sol.**

In BCF and CBE, we have

[Since Each = 90Âº]

Hyp. BC = Hyp. BC [Common]

FC = EB

Therefore By R.H.S. criterion of congruence, we have

[Since corresponding parts of congruent triangles are equal]

Now, in ABC

[Since ]

AB = AC

[Since sides opposite to equal angles of a triangle are equal]

Therefore is an isosceles triangle.

**Q.5Â Â Â Â Â ABC is an isosceles triangle with AB = AC. Draw AP BC to show that **

**Sol.**

In ABP and ACP, we have

AB = AC [Given]

AP = AP [Common]

and [Since each = 90Âº]

Therefore , by R.H.S. criterion of congruence we have

[Since corresponding parts of congruent triangles are equal]