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 $PQ \parallel AB$

$\angle ABP =\angle BPQ$.........…(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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Updated on: 10-Oct-2022

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