In given figure, at common base BC, two isosceles triangles ∆PBC and

1.	In given figure, at common base BC, two isosceles triangles ∆PBC and ∆QBC lie on both sides of BC. Prove that line joining P and Q bisects line BC at 90°.
Answer» Given : ∆PBC and ∆QBC are two isosceles triangles which lie on both sides of base BC. Here, PB = PC and BQ = CQ To Prove : ∠POB = ∠POC = 90° or ∠QOB = ∠QOC = 90° Proof : In ∆PBC, PB = PC (Given) ∴ ∠PBO = ∠PCO (Equal sides) PO = PO (Common) by S.A.S. congruence of ∆POB ≅ ∆POC (Corresponding sides of corresponding triangles are equal) ⇒ ∠PBO = ∠POC ….(i) We Know that ∠PBO + ∠POC = 180° ∠PBO + ∠POB = 180° [From equation (i)] 2∠POB = 180° ∠POB = 180°/2 = 90° ∠PBO = ∠POC = 90° Similarly, ∠QOB = ∠QOC = 90° Thus, PQ, bisects BC at 90°

In given figure, at common base BC, two isosceles triangles ∆PBC and ∆QBC lie on both sides of BC. Prove that line joining P and Q bisects line BC at 90°.

Answer»

Given : ∆PBC and ∆QBC are two isosceles triangles which lie on both sides of base BC.

Here, PB = PC and BQ = CQ

To Prove : ∠POB = ∠POC = 90°

or ∠QOB = ∠QOC = 90°

Proof : In ∆PBC,

PB = PC (Given)

∴ ∠PBO = ∠PCO (Equal sides)

PO = PO (Common)

by S.A.S. congruence of

∆POB ≅ ∆POC (Corresponding sides of corresponding triangles are equal)

⇒ ∠PBO = ∠POC ….(i)

We Know that

∠PBO + ∠POC = 180°

∠PBO + ∠POB = 180° [From equation (i)]

2∠POB = 180°

∠POB = 180°/2 = 90°

∠PBO = ∠POC = 90°

Similarly, ∠QOB = ∠QOC = 90°

Thus, PQ, bisects BC at 90°