CBSE Class 9 Answered

The question must be:

PR is the bisector of ∠QPS and PT is perpendicular to QS. Prove that ∠TPR = half[∠Q - ∠S]

Given: PR is bisector of ∠QPS
PT is perpendicular to QS

To Prove: ∠TPR = 1/2 (∠Q - ∠S)

Proof:

We know that ∠QPR = 1/2 ∠P .... (i)

∠Q + ∠QPT = 90^o 

∠QPT = 90^o - ∠Q  .... (ii)

We know that

∠TPR = ∠QPR - ∠QPT
= 1/2 ∠P - (90^o - ∠Q)        ... From (i) and (ii)
= 1/2 ∠P - 90^o + ∠Q
= 1/2 ∠P - 1/2 (∠Q + ∠P + ∠S) + ∠Q
= 1/2 ∠P - 1/2 ∠Q - 1/2 ∠P - 1/2 ∠S + ∠Q
= -1/2 ∠Q - 1/2 ∠S + ∠Q (+ve and -ve 1/2 ∠P got cancelled)

By takig LCM in ∠Q we get

∠TPR = ∠QPR - ∠QPT
= -1/2 ∠Q + 2/2 ∠Q - 1/2 ∠S
= (-1/2 + 2/2) ∠Q - 1/2 ∠S
= 1/2 ∠Q - 1/2 ∠S

∠TPR = 1/2 ( ∠Q - 1/2 ∠S )

Hence, proved.