Question

PS is the bisector of ∠QPR, Prove that
1) ∠QPS+∠PSR= ∠PRT
2) ∠PQS+∠PRT= 2 ∠PSR

Answer 1

Answer:

Step-by-step explanation:

PR > PQ,

∴ ∠PQR > ∠PRQ (Angles opp. to the longer side is greater)

PS is the bisector of ∠QPR.

∴ ∠QPS = ∠RPS

Let ∠QPS = ∠RPS = y

In ΔPQS,

∠RPS is the exterior angle.

We know that exterior angle is sum of interior opp. angles

∴ ∠PSR = ∠PQR + y   …. (1)

In ΔPSR,

∠PSQ is the exterior angle.

∴ ∠PSQ = ∠PRQ + y   …. (2)

Now,

∠PQR > ∠PRQ

Adding y on both sides to get,

∠PQR + y> ∠PRQ + y

∠PSR > ∠PSQ

Hence Proved.

Answer 2

PR > PQ,

》∠PQR > ∠PRQ (Angles opp. to the longer side is greater)

PS is the bisector of ∠QPR.

》 ∠QPS = ∠RPS

♧Let ∠QPS = ∠RPS = y♧

In ΔPQS,

∠RPS is the exterior angle.

We know that exterior angle is sum of interior opp. angles

》∠PSR = ∠PQR + y   …. (1)

In ΔPSR,

∠PSQ is the exterior angle.

》 ∠PSQ = ∠PRQ + y   …. (2)

Now,

∠PQR > ∠PRQ

Adding y on both sides to get,

∠PQR + y> ∠PRQ + y

∠PQR + y> ∠PRQ + y∠PSR > ∠PSQ

Hence Proved.