please solve as fast possible
Attachments:
Answers
Answered by
1
Answer:
∠PAD
= ∠PAB + ∠BAD
= ∠ACB + ∠BAD [ tangent - secant theorem ]
= ∠ACB + ∠CAD [ angle bisector ]
= ∠PDA [ external angle of a triangle ]
Therefore ΔPAD is isosceles.
Similar questions