Question

please solve the questiion friends...

Answer 1

Step-by-step explanation:

1. in ∆BPC and ∆EQD,

BC = ED (given)

we know that the exterior angle is equal to the sum of two interior angles

so, (angel BCD = angel CPB + angel PBC) =

( angel EDC = angel DQE + angel QED)

thus ∆ BPC ~ ∆ EQD

so, BP = EQ

and PC = QD

Hence the sides of ∆ BPC = the sides of ∆ EQD

2. since ∆ BPC ~ ∆ EQD

PB = QE (1)

now in ∆ APQ we have,

PB + BA = QE + EA { using (1) }

AP = AQ ( proved)

I hope this answer will help you in our math problems