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In the figure, $AB \parallel CD$ and $P$ is any point shown in the figure. Prove that:$\angle ABP + \angle BPD + \angle CDP = 360^o$"


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 $CD$


$AB \parallel PQ$

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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Updated on: 10-Oct-2022

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