Question

In the figure, $AD\perp CDandCB\perp CD.IfAQ=BPandDP=CQ,\text{prove that}\angle DAQ=\angle CBP.$

Answer 1

Given : In the figure,

$AD\perp CDandCB\perp CD,AQ=BPandDP=CQ$

To prove : $\angle DAQ=\angle CBP$

Proof : $\because DP=CQ$

$\therefore DP+PQ=PQ+QC$

$\Rightarrow DQ=PC$

Now in right $\Delta ADQand\Delta BCP$

Side DQ = PC (Proved)

Hyp. AQ = BP

$\therefore \Delta ADQ\cong \Delta BCP\text{(RHS axiom)}$

$\therefore \angle DAQ=\angle CBP(c.p.c.t.)$