Question

Que: In the Figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.​

Answer 1

Solution:

Given, PR > PQ and PS bisects ∠QPR

To prove: ∠PSR > ∠PSQ

Proof:

∠QPS = ∠RPS — (1) (PS bisects ∠QPR)

∠PQR > ∠PRQ — (2) (Since PR > PQ as angle opposite to the larger side is always larger)

∠PSR = ∠PQR + ∠QPS — (3) (Since the exterior angle of a triangle equals the sum of opposite interior angles)

∠PSQ = ∠PRQ + ∠RPS — (4) (As the exterior angle of a triangle equals to the sum of opposite interior angles)

By adding (1) and (2)

∠PQR + ∠QPS > ∠PRQ + ∠RPS

Now, from (1), (2), (3) and (4), we get

∠PSR > ∠PSQ

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Hope it will be helpful :)

Answer 2

Solution:-

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In ∆ PQR

PR > PQ ( Given )

.°. < PQR > <PRQ ---------(1)

( A triangles longest sides of opposite angle is big )

°.° PS , < QPRs a bisect.

.°. < QPS = < RPS

in ∆ PQR ------------(2)

< PQR + < QPS + < PSQ = 180° ----------(3)

( all angles of a triangle equal is 180° )

In ∆ PRS

< PRS + < SPR + < PSR = 180° -------(4)

( all angles of a triangle equal is 180° )

From (3) and (4) ,

< PQR + < QPS + < PSQ = < PRS + < SPR + < PSR

=> < PQR + < PSQ = < PRS + < PSR

=> < PRS + < PSR = < PQR + < PSQ

=> < PRS + < PSR > < PRQ + < PSQ ( From (1) )

=> < PRQ + < PSR > < PRS + < PSQ ( °.° < PRQ = < PRS

=> < PSR > < PSQ

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