Goutham wrote: Let $O$ be the circumcenter of triangle $ABC$. A line through $O$ intersects the sides $CA$ and $CB$ at points $D$ and $E$ respectively, and meets the circumcircle again at point $P \neq O$ inside the triangle. A point $Q$ on side $AB$ is such that $\frac{AQ}{QB}=\frac{DP}{PE}$. Prove that $\angle APQ = 2\angle CAP$. Proposed by Dusan Djukic Sorry but since the circumcircle lies outside $\triangle ABC,$ a line can never meet it at any point inside the triangle... Can you please correct the typo and post the correct version? Thanks in advance.

Quote: and meets the circumcircle again at point Circumcircle of triangle $ABO$.

Reflect $P$ in $AC, BC$ into $P_a, P_b.$ $\angle P_aAB + \angle ABP_b = 2 (\angle CAP + \angle PBC) + (\angle PAB + \angle ABP) =$ $2 \angle BCA + 180^\circ - 2 \angle BCA = 180^\circ$ $\Longrightarrow$ $AP_a \parallel BP_b.$ Let $OP$ cut $AB$ at $S$ $\Longrightarrow$ $\frac{SA}{SB} = \frac{PA}{PB}$ $\Longrightarrow$ $S$ is external similarity center of circles $(A), (B)$ with radii $AP = AP_a, BP = BP_b$ $\Longrightarrow$ $ S \in P_aP_b.$ Let $OP$ cut $AP_a, BP_b$ at $U, V$ $\Longrightarrow$ $\frac{AU}{AP_a} = \frac{BV}{BP_b}$ $\Longrightarrow$ $\frac{PD}{PU} = \frac{PE}{PV}$ $\Longrightarrow$ $\frac{PU}{PV} = \frac{PD}{PE} = \frac{QA}{QB}$ $\Longrightarrow$ $QP \parallel AUP_a \parallel BVP_b$ $\Longrightarrow$ $\angle APQ = \angle P_aAP =2 \angle CAP.$

Let $R$ be the point on $AB$ such that $\angle APR=2\angle CAP$. Also let $\angle CAP=\alpha$, $\angle CBP=\beta$, $\angle PAB=x,\angle PBA=y$. By some angle chasing we easily get $\angle APR=2\alpha$, $\angle BPR=2\beta$, $\angle APD=\angle BPE=90^{\circ}-\alpha-\beta$, $\angle ADP=90^{\circ}+\beta$ and $\angle BPE=90^{\circ}+\alpha$. Now we have \[\frac{DP}{PE}=\frac{DP}{PA}\cdot\frac{PA}{PB}\cdot\frac{PB}{PE}=\frac{\sin\alpha}{\sin(90^{\circ}+\beta)}\cdot\frac{\sin y}{\sin x}\cdot\frac{\sin(90^{\circ}+\alpha)}{\sin\beta}=\frac{\sin\alpha\cos\alpha\sin y}{\sin\beta\cos\beta\sin x}=\frac{\sin2\alpha}{\sin x}\cdot\frac{\sin y}{\sin2\beta}=\frac{AR}{PR}\cdot\frac{PR}{RB}=\frac{AR}{RB}\] Therefore $\frac{AR}{RB}=\frac{AQ}{QB}$, which implies that $R\equiv Q$. Since $\angle APR=2\angle CAP$, then we are done.

draw $EK$ parallel to $AC$,$K$ is on line $AP$,$\angle EPK=\angle BPE=90-\angle C$ $\angle BEK=180-\angle C$,so $E$ is the incenter of triangle $BPK$, so $\angle AKB=2\angle CAP$

Let circle around $\triangle CDE$ intersect the circumcircle at $X$ and let $DE$ intersect $AB$ at $P'$ and let $C'$ be opposite to $O$ . $X$ is the center of spiral similarity that maps $ D-P-E \rightarrow A-Q-B$ so $XP'AD$ is cyclic so $\angle XP'O=\angle XAC=\angle XC'O$ so $XP'OC'$ is cyclic. Now when we invert around circumcircle $P'$ goes to $P$ and $X$ and $C'$ stay fixed so $X,P,C'$ are colinear. Now by easy angle chase $\angle DXC=90+\angle DAP$ so $\angle DXP=\angle DAP$. Now let QP intersect $AC$ at $A'$. Now because $X$ maps $DP$ to $AQ$ we have $AQA'X$ and $DPA'X$ are cyclic so $\angle PA'A=\angle PAA'$ so $\angle APQ=2\angle CAP$ so we are finished.