Question

DPQ is an isosceles triangle with DP = DQ. A straight line CD bisects the exterior ∠QDR. Prove that DC is parallel to PQ.

Answer 1

In ΔQDP,
DP = DQ
∴ ∠Q = ∠P
∠QDR = ∠Q + ∠P
2∠QDC = ∠Q + ∠P      ....(DC bisects angle QDR)
2∠QDC = ∠Q + ∠Q = 2∠Q
∠QDC = ∠Q
But these are alternate angles.
∴ DC || PQ.