## Chapter 5 Triangles Ex 5.3

**Question 1.** **∆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**

**(i) ∆ABD = ∆ACD** **(ii) ∆ABP = ∆ACP** **(iii) AP bisects ∠A as well as ∠D** **(iv) AP is the perpendicular bisector of BC.** **Solution:Given** ∆ABC and ∆DBC are two isosceles triangles having common

base BC, suchthat AB=AC and DB=OC.

**To prove:**

(i) ∆ABD = ∆ACD

(ii) ∆ABP = ∆ACP

(iii) AP bisects ∠A as well as ∠D

(iv) 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 ∠A** **Solution:**

In ∆ ABD and ∆ ACD, we have

AB = AC (Given)

∠ADB = ∠ADC = 90° (∵ Given AD ⊥BC)

AD = AD (Common)

∴ ∆ ABD ≅ ∆ ACD (By RHS congruence axiom)

BD=DC (By CPCT)

⇒ AD bisects BC.

∠ BAD = ∠ CAD (By CPCT)

∴ AD bisects ∠A .

**Question 3.**
**Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that**
**(i) ∆ABC ≅ ∆PQR**
**(ii) ∆ABM ≅ ∆PQN**

**Solution:**

**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 ∆BEC and ∆CFB, we have

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

In ∆ABP and ∆ACP, We have

AB = AC (Given)

AP = AP (Common)

and ∠APB = ∠APC = 90° (∵ AP ⊥ BC)

∴ ∆ABP ≅ ∆ACP (By RHS congruence axiom)

⇒ ∠B = ∠C (By CPCT)