Question

Two isosceles triangle PQR and PQS are drawn on the opposite sides of a common base PQ. if

Answer 1

Answer:

Now in ∆ PRS and ∆ PQS

PR = PQ ( Given )

SR = QS ( Given )

PS = PS ( Common side )

Hence,

∆ PRS ≅ ∆ PQS ( By SSS rule )

so,

∠ QPO = ∠ RPO ( By CPCT )

Here PQR is a isosceles triangle as PQ = PR

So,

∠ PQO = ∠ PRO ( by base angle theorem )

In ∆ PQO and ∆ PRO

∠ QPO = ∠ RPO ( As we proved )

PQ = PR ( Given )

∠ PQO = ∠ PRO ( by base angle theorem )

Hence,

∆ PQO ≅ ∆ PRO

So,

OQ = OR ------- ( 1 ) ( By CPCT )

And

∠ POQ = ∠ POR ------- ( 2 ) ( By CPCT )

And

∠ POQ + ∠ POR = 180° ( Linear pairs are supplementary, definition of supplementary angles )

from equation 2 we get

∠ POQ + ∠ POQ = 180°

2 ∠ POQ = 180°

∠ POQ = 90°

So.

∠ POQ = ∠ POR = 90° ------------ ( 3 )

So from equation 1 and equation 3 we get

PS bisects QR at right angle. ( Hence proved )