Question

how to do it its urgent

Answer 1

$\red{\bold{\underline{Answer \:for\: your\:question}}}$

$\red{\underline{Given\colon}}$

(a) AD & CB perpendicular to CD

(b) AQ = BP

(c) DP = CQ

$\red{\underline{To\: Prove \colon}}$

$\angle{DAQ} = \angle{CBP}$

$\red{\underline{Solution}}$

$Add \: PQ \: both \: side\: in \: eq(c)\\\\ DP\: + \: PQ = PQ \: + \: QC \\\\ \bold{\boxed{ DQ = PC }} --(eq \:\:i )$

$\: Now \: in\: \triangle{ADQ}\:\: \&\: \: \triangle{BPC}$

[1] AQ = BP ......{ given }

$[2] \:\angle{ADQ} = \angle{BCP}$ ......{90° each }

[3] DQ = PC ......{eq i}

The three pairs of triangles are equal one is Right Angle ,Hypotenuse , and One Side.

$\red{\bold{\underline{RHS\: Congruency \: Rule }}}$

∆ ADQ congruent to ∆ BPC

Congruency rule ( RHS )

Now ,

$\blue{BY \: C.P.C.T.}$

$\angle{DAQ} = \angle{CBP}$

$\red{\underline{Answer}}$

$\boxed{\angle{DAQ} = \angle{CBP}}$