Question

In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that $\angle ABP+\angle CDP=\angle DPB$.

Answer 1

Given: In figure, AB ||CD P is a point between AB and CD PD and PB are joined
To prove: $\angle APB+\angle CDP=\angle DPB$
Construction: Through P, draw PQ ||AB or CD
Proof: $\because AB||PQ$
$\therefore \angle ABP=BPQ$ ...(i)
(Alternate angles)
Similarly,
CD||PQ
$\therefore \angle CDP=DPQ$ ...(ii)
(Alternate angles)
Adding (i) and (ii)
$\angle ABP+\angle CDP=\angle BPQ+\angle DPQ$
Hence $\angle ABP+\angle CDP=\angle DPB$