In the figure, $AD \perp CD$ and $CB \perp CD$. If $AQ = BP$ and $DP = CQ$, prove that $\angle DAQ = \angle CBP$.

Given:

$AD \perp CD$ and $CB \perp CD$.

$AQ = BP$ and $DP = CQ$.

To do:

We have to prove that $\angle DAQ = \angle CBP$.

Solution:

$DP = CQ$

This implies,

$DP + PQ = PQ + CQ$

$DQ = PC$

In $\triangle ADQ$ and $\triangle BCP$

$DQ = PC$

$AQ = BP$

Therefore, by RHS axiom,

$\triangle ADQ \cong \triangle BCP$

This implies,

$\angle DAQ = \angle CBP$           (CPCT)

Hence proved.

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Updated on: 10-Oct-2022

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