What is $m(\overarc{PAC})$?

Hint

Best regards, sunken rock

we have two cases: 1. The point $P$ lies on the segment $BQ$. Then $ \angle BAP+\angle BCQ=180^\circ $. 2. The point $B$ lies on the segment $PQ$. Then $ \angle BAP=\angle BQC$.

lyukhson wrote: What is $m(\overarc{PAC})$? Oh, it means $\angle PAC$ we have slightly different notations in Turkey.

Ceva theorem inside and outside. Done. Pray for Malaysian flight MH370!...

how do we prove the fact that $BR$ is parallel to $CP$? @VoDucDien: Which triangles and points do we do the Ceva's Theorem to?

I have a nice solution: Let $X=AP \cap CQ$ and $J=QD \cap BX$.Assume that $BX$ is nonparallel with $CP$ and $K$ be the intersection point of $CP$ and $BX$.Note that $XP,BC,QJ$ are concurrent cevians in $\triangle{BXQ}$ so $(C,P;D,K)=-1$.Seeing this from $J$ we get that $JC,JP,JD,JK$ form a harmonic pencil.Now since $JD$ is the median we must have $JK \parallel PC$.This is a contradiction to our assumption.Hence $BX \parallel CP$. The rest is trivial:Note that $\angle{XAC}=\angle{BCP}=\angle{XBC}$ so $ABXC$ is cyclic.Thus $\angle{BCQ}=\angle{BCP}+\angle{PCQ}=\angle{BCP}+\angle{BXC}=\angle{BCP}+180^{\circ}-\angle{BAC}=180^{\circ}-(\angle{BAC}-\angle{EAC})=180^{\circ}-\angle{BAE} \implies \sin{\angle{BAE}}=\sin{\angle{BCQ}}$ as desired.

I have a short proven : due sucken rock we call R is intersect of AP and CQ , we must prove CP//BR , we can see the confiration of trapezium lemma . assume CP intersect BR at X we have Q(XDPC)=-1 and DP=DC==> QX//PC (contract) WAD edit : it is similar above

emregirgin35 wrote: Let $P$ be a point inside the acute triangle $ABC$ with $m(\widehat{PAC})=m(\widehat{PCB})$. $D$ is the midpoint of the segment $PC$. $AP$ and $BC$ intersect at $E$, and $BP$ and $DE$ intersect at $Q$. Prove that $\sin\widehat{BCQ}=\sin\widehat{BAP}$. Assume that $AP$ intersects the circumcircle of $ABC$ at $T$ for the second time. We have $\angle CBT=\angle CAT=\angle PCB\Rightarrow BT//PC$. Since $DE$ bisects $[PC]$ and $BT//PC$, we find that $DE$ also bisects $[BT]$. Now, see that these imply $Q, C, T$ are collinear. Hence, $\angle ECQ=180^\circ-\angle BCT=\angle 180^\circ-\angle BAP$, as desired.