## Triangles

**Chapter 7**

**Exercise 7.3**

**EX 7.3 QUESTION 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:**

(i) In ∆ABD and ∆ACD,

AB = AC [Given]

AD = DA [Common]

BD = CD [Given]

∴ ∆ABD ≅ ∆ACD [By SSS congruency]

∠BAD = ∠CAD [By C.P.C.T.] …(1)

(ii) In ∆ABP and ∆ACP,

AB = AC [Given]

∠BAP = ∠CAP [From (1)]

∴ AP = PA [Common]

∴ ∆ABP ≅ ∆ACP [By SAS congruency]

(iii) Since, ∆ABP ≅ ∆ACP

⇒ ∠BAP = ∠CAP [By C.P.C.T.]

∴ AP is the bisector of ∠A.

Again, in ∆BDP and ∆CDP,

we have BD = CD [Given]

DP = PD [Common]

BP = CP [ ∵ ∆ABP ≅ ∆ACP]

⇒ A BDP = ACDP [By SSS congruency]

∴ ∠BDP = ∠CDP [By C.P.C.T.]

⇒ DP (or AP) is the bisector of ∠BDC

∴ AP is the bisector of ∠A as well as ∠D.

(iv) ∆ABP ≅ ∆ACP

⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.]

But ∠APB + ∠APC = 180° [Linear Pair]

∴ ∠APB = ∠APC = 90°

⇒ AP ⊥ BC, also BP = CP

∴ AP is the perpendicular bisector of BC.

**EX 7.3 QUESTION 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:**

(i) In ∆ABD and ∆ACD

AB =AC [Given]

∠ADB = ∠ADC [Each 90°]

AD = DA [Common]

∴ ∆ABD ≅ ∆ACD [By RHS congruency]

So, BD = CD [By C.P.C.T.]

⇒ D is the mid-point of BC or AD bisects BC.

(ii) ∠BAD = ∠CAD [By C.P.C.T.]

So, AD bisects ∠A

**EX 7.3 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 ΔPQR (see Fig. 7.40). Show that:**

**(i) ΔABM ΔPQN**

**(ii) ΔABC ΔPQR**

**Solution:**

BC = QR [Given]

½BC = ½QR

⇒ BM = QN [AM and PN are median.]

(i) In ∆ABM and ∆PQN,

AB = PQ , [Given]

AM = PN [Given]

BM = QN [proved above]

∴ ∆ABM ≅ ∆PQN [By SSS congruency]

(ii) In ∆ABM ≅ ∆PQN [proved above]

⇒ ∠B = ∠Q [By C.P.C.T.]

Now, in ∆ABC and ∆PQR

∠B = ∠Q [proved above]

AB = PQ [Given]

BC = QR [Given]

∴ ∆ABC ≅ ∆PQR [By SAS congruency]

**EX 7.3 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:**

In ΔECB and ΔFBC,

∠BFC = ∠CBE [Each 90°]

BC = CB [Common]

FC = BE [Given]

∆ECB ≅ ∆FBC [ RHS congruency]

∠FBC = ∠ECB [C.P.C.T.]

⇒ AB = AC [Sides opposite to equal angles of a ∆ are equal]

∴ ABC is an isosceles triangle.

**EX 7.3 QUESTION 5.**

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

**Solution:**

In ∆ABP and ∆ACP,

∠APB = ∠APC [Each 90°]

AB = AC [Given]

AP = AP [Common]

∴ ∆ABP ≅ ∆ACP [By RHS congruency]

So, ∠B = ∠C [By C.P.C.T.]