**Exercise: 7.3 (Page No: 128)**

**1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that**

(i) ΔABD ΔACD

(ii) ΔABP ΔACP

(iii) AP bisects A as well as D.

(iv) AP is the perpendicular bisector of BC.

**Solution:**

In the above question, it is given that ΔABC and ΔDBC are two isosceles triangles.

(i) ΔABD and ΔACD are similar by SSS congruency because:

AD = AD (It is the common arm)

AB = AC (Since ΔABC is isosceles)

BD = CD (Since ΔDBC is isosceles)

∴ ΔABD ΔACD.

(ii) ΔABP and ΔACP are similar as:

AP = AP (It is the common side)

PAB = PAC (by CPCT since ΔABD ΔACD)

AB = AC (Since ΔABC is isosceles)

So, ΔABP ΔACP by SAS congruency condition.

(iii) PAB = PAC by CPCT as ΔABD ΔACD.

AP bisects A. — (i)

Also, ΔBPD and ΔCPD are similar by SSS congruency as

PD = PD (It is the common side)

BD = CD (Since ΔDBC is isosceles.)

BP = CP (by CPCT as ΔABP ΔACP)

So, ΔBPD ΔCPD.

Thus, BDP = CDP by CPCT. — (ii)

Now by comparing (i) and (ii) it can be said that AP bisects A as well as D.

(iv) BPD = CPD (by CPCT as ΔBPD ΔCPD)

and BP = CP — (i)

also,

BPD +CPD = 180° (Since BC is a straight line.)

⇒ 2BPD = 180°

⇒ BPD = 90° —(ii)

Now, from equations (i) and (ii), it can be said that

AP is the perpendicular bisector of BC.

**2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that**

(i) AD bisects BC (ii) AD bisects A.

**Solution:** It is given that AD is an altitude and AB = AC. The diagram is as follows:

(i) In ΔABD and ΔACD,

ADB = ADC = 90°

AB = AC (It is given in the question)

AD = AD (Common arm)

∴ ΔABD ΔACD by RHS congruence condition.

Now, by the rule of CPCT,

BD = CD.

So, AD bisects BC

(ii) Again, by the rule of CPCT, BAD = CAD

Hence, AD bisects A.

**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 ΔPQR (see Fig. 7.40). Show that:**

(i) ΔABM ΔPQN

(ii) ΔABC ΔPQR

**Solution:**

Given parameters are:

AB = PQ,

BC = QR and

AM = PN

(i) ½ BC = BM and ½ QR = QN (Since AM and PN are medians)

Also, BC = QR

So, ½ BC = ½ QR

⇒ BM = QN

In ΔABM and ΔPQN,

AM = PN and AB = PQ (As given in the question)

BM = QN (Already proved)

∴ ΔABM ΔPQN by SSS congruency.

(ii) In ΔABC and ΔPQR,

AB = PQ and BC = QR (As given in the question)

ABC = PQR (by CPCT)

So, ΔABC ΔPQR by SAS congruency.

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

**Solution:**

It is known that BE and CF are two equal altitudes.

Now, in ΔBEC and ΔCFB,

BEC = CFB = 90° (Same Altitudes)

BC = CB (Common side)

BE = CF (Common side)

So, ΔBEC ΔCFB by RHS congruence criterion.

Also, C = B (by CPCT)

Therefore, AB = AC as sides opposite to the equal angles is always equal.

**5. ABC is an isosceles triangle with AB = AC. Draw AP ****⊥**** BC to show that B = C.**

**Solution:**

In the question, it is given that AB = AC

Now, ΔABP and ΔACP are similar by RHS congruency as

APB = APC = 90° (AP is altitude)

AB = AC (Given in the question)

AP = AP (Common side)

So, ΔABP ΔACP.

∴ B = C (by CPCT)