In the figure, $AB \parallel CD$ and $P$ is any point shown in the figure. Prove that:
$\angle ABP + \angle BPD + \angle CDP = 360^o$
![](/assets/questions/media/153848-1631286879.png)
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Given:
In the figure, $AB \parallel CD$ and $P$ is any point shown in the figure.
To do:
We have to prove that $\angle ABP + \angle BPD + \angle CDP = 360^o$.
Solution:
Through $P$, draw $PQ \parallel AB$ and $CD$
![](/assets/questions/media/153848-52584-1631370827.png)
$AB \parallel PQ$
This implies,
$\angle ABP + \angle BPQ= 180^o$....……(i) (Sum of co-interior angles is $180^o$)
Similarly,
$CD \parallel PQ$
This implies,
$\angle QPD + \angle CDP = 180^o$......…(ii)
Adding equations (i) and (ii), we get,
$\angle ABP + \angle BPQ + \angle QPD + \angle CDP = 180^o+ 180^o$
$ = 360^o$
$\Rightarrow \angle ABP + \angle BPD + \angle CDP = 360^o$
Hence proved.
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