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2. Let two circles $C_1$ and $C_2$ intersect in points $A$ and $B$. The tangent to $C_1$ at $A$ intersects $C_2$ in $P$ and the line $PB$ intersects $C_1$ for the second time in $Q$ (suppose that $Q$ is outside $C_2$). The tangent to $C_2$ from $Q$ intersects $C_1$ and $C_2$ in $C$ and $D$, respectively (The points $A$ and $D$ lie on different sides of the line $PQ$). Show that $AD$ is bisector of the angle $CAP$.

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. $\angle CAD=\angle PAD$

$[Done]$