In the figure, lines $AB$ and $CD$ are parallel and $P$ is any point as shown in the figure. Show that $\angle ABP + \angle CDP = \angle DPB$.
"
Given:
Lines $AB$ and $CD$ are parallel and $P$ is any point as shown in the figure.
To do:
We have to show that $\angle ABP + \angle CDP = \angle DPB$.
Solution:
Through $P$, draw $PQ \parallel AB$
![](/assets/questions/media/153848-52583-1631370417.png)
$\angle ABP =\angle BPQ$.........…(i) (Alternate angles)
Similarly,
$CD \parallel PQ$
$\angle CDP = \angle DPQ$.....…(ii) (Alternate angles)
Adding equations (i) and (ii), we get,
$\angle ABP + \angle CDP = \angle BPQ + \angle DPQ$
Hence, $\angle ABP + \angle CDP =\angle DPB$.
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