Question

the sides AB , AC of triangle ABC are equal and P is any point within the triangle on the bisector of angle BAC . BP produced meets AC in Q .prove that BP> PQ

Answer 1

Given: AB = AC
angle BAP = angle CAP
To Prove: BP >PQ

Construction : Join PC

Proof : In Δ APB and APC

AB =AC
angle BAP= angle CAP
AP= Common side
Δ APB congruent to Δ APC

therefore , BP= CP（c.p.c.t）
angle ABP= Angle ACP

In Δ ABQ,
Angle PQC is the exterior angle.
angle PQC > angle ABP
so,anglePQC>AngleACP（ ABP=ACP）
therefore,CP> PQ
BP > PQ（as ,BP= CP）
proved

Answer 2

Step-by-step explanation:

hence proved

see the solution in photo

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