Question

Q. 6. 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 bi-
sects line BC at 90°.​

Answer 1

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 90Read more on Sarthaks.com - https://www.sarthaks.com/757817/in-given-figure-at-common-base-bc-two-isosceles-triangles-pbc-and-qbc-lie-on-both-sides-of

Answer 2

Answer:

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°Read more on Sarthaks.com - https://www.sarthaks.com/757817/in-given-figure-at-common-base-bc-two-isosceles-triangles-pbc-and-qbc-lie-on-both-sides-of