Question

In the figure, AB || CD and P is any point shown in the figure. Prove $\angle ABP+\angle BPD+\angle CDP={360}^0$

Answer 1

Given: In figure, AB ||CD P is any point as shown in the figure
To prove: $\angle ABP+\angle BPD+\angle CDP={360}^0$
Construction: Through P, draw PQ ||AB or CD
Proof: $\because AB||PQ$
$\therefore \angle ABP+BPQ={180}^0$ ...(i)
(Sum of co-interior angles)
Similarly, CD||PQ
$\therefore \angle QPD+\angle CDP={180}^0$ ...(ii)
Adding (i) and (ii)
$\angle ABP+\angle BPQ+\angle QPD+\angle CDP={180}^0+{180}^0={360}^0$
$\Rightarrow \angle ABP+\angle BPD+\angle CDP={360}^0$