Hunting In December.

Let $D$ be a point on $\overline{\rm BC}$ such that $\overline{\rm CD}$ $=$ $\overline{\rm CA}$ . Draw $\overline{\rm AD}$ , $\overline{\rm AI}$ , $\overline{\rm BI}$ & $\overline{\rm CI}$ .

$\overline{\rm BD}$ $=$ $\overline{\rm BC}$ $-$ $\overline{\rm CD}$ $=$ $\overline{\rm BC}$ $-$ $\overline{\rm AC}$ $=$ $\overline{\rm AI}$

$\angle {ADB}$ $=$ $\angle {ACD}$ $+$ $\angle {DAC}$ $=$ $\angle {ACD}$ $+$ $90$ $-$ $\frac{\angle {ACD}}{2}$ $=$ $90$ $+$ $\frac{\angle {ACD}}{2}$ $=$ $\angle {AIB}$

In triangles $\bigtriangleup AIB$ and $\bigtriangleup BDA$ , $\overline{\rm AI}$ $=$ $\overline{\rm BD}$ , $\angle {AIB}$ $=$ $\angle {ADB}$ & $\overline{\rm AB}$ is common to them. Therefore, $\bigtriangleup AIB$ $\cong$ $\bigtriangleup BDA$ by the $S-S-A$ criterion of congruence.

Hence, $\angle {BAI}$ $=$ $\angle {ABD}$ $=$ $35^{\circ}$ $\Rightarrow$ $\angle {BAC}$ $=$ $2$ $\angle {BAI}$ $=$ $70$

$\angle {ACB}$ $=$ $180-70-35$ $=$ $75^{\circ}$

Let $O$ be the circumcentre of $\bigtriangleup ABC$ . Draw $\overline{\rm OB}$ and $\overline{\rm OC}$ . $\overline{\rm OB}$ $\perp$ $\overline{\rm BP}$ and $\overline{\rm OC}$ $\perp$ $\overline{\rm CP}$ . $\Rightarrow$ $BOCP$ is a cyclic quadrilateral.

$\bigtriangleup BPC$ is isosceles. $\Rightarrow$ $\angle {BCP}$ $=$ $\frac{180-\angle {BPC}}{2}$ $=$ $\frac{\angle {BOC}}{2}$ $=$ $\angle {BAC}$ .

Observe that, $\angle {BQP}$ $=$ $\angle {BAC}$ $=$ $\angle {BCP}$ , which points out that $BQCP$ is a cyclic quadrilateral.

Hence, $\angle {AQC}$ $=$ $180-\angle{BQC}$ $=$ $\angle {BPC}$ $=$ $180-\angle{BOC}$ $=$ $180-2\angle{BAC}$ $=$ $180-2{\times}70$ $=$ $\boxed{\textbf{40}}$ .