Question

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

Answer 1

Given: $AD\perp CD,CB\perp CD,AQ=BPandDP=CQ$

Since, DP = QC

Adding PQ on both sides,we get

DP + PQ = PQ + QC

$\Rightarrow DQ=PC....\left(1\right)$

In $△ADQ$ and $△BCP$,

DQ = PC (from 1)

AQ = BP (Given)

$\angle ADQ=\angle BCP\left({90}^{\circ }each\right)$

$\therefore △ADQ\cong △BCP\left(byRHSaxiom\right)$

Thus, $\angle DAQ=\angle CBP(byC.P.C.T)$

Hence, proved.