Post Number:#2 by Thamim Zahin » Thu Jan 12, 2017 3:45 pm
Extend the segment $QD$. Name it $C'$.
$\angle CQB=\angle BAC=\angle a$
$\angle BAD=\angle DPB=\angle b$ [cyclic quad]
$\angle DAP=\angle C'DP=\angle a$ [$DC'$ is tangent to $(DAP)$]
Now, notice that $\angle DAC=\angle b+\angle c$ and $\angle DAP=\angle a$
So, we have to proof that $\angle c+\angle b=\angle a$
Now, in triangle $\triangle DPQ$, $DC'$ is extension of $QD$.
So, $\angle c+\angle b=\angle a$
Or. $\angle CAD=\angle PAD$
$[Done]$
 Attachments

 I2.png (48.01 KiB) Viewed 48 times