China owns the license to weird problems. Let \(\odot BDC \cap AB = F\) and \(\odot CDE \cap AB = G\), in \(\angle DFA = \angle C, \angle DGE = \angle DCE = \angle C/2 \Rightarrow \Delta DFG\) is isosceles and \(DF = FG\) but due to cyclic quadrilateral \(\square BCDF, CD = DF\) so, \(CD=DF=FG.\) Now, note that \(\Delta ADP \cong \Delta BGC\) (ASA - congruency axiom). \(\Rightarrow CG = PD \Rightarrow \Delta BFC \cong \Delta APC \Rightarrow \angle BCF = \angle ACP \Rightarrow \measuredangle {[BC,PC]} = \measuredangle {B/2} \quad \square\)

Let BD∩CE=I, Due to ED/DP=AE*sin∠BAC/AP*sin∠PAC=(AE*sin∠BAC)/(AP*sin∠ABC)=(AE*BC)/(AP*AC)=AE/AC=EI/IC so,BD∥CP

When we're talking about bisectors, are we talking about length bisector or angle bisector

"the bisectors of $\angle ABC, \angle ACB$" so it's about angles

Let $BC=a,CA=b,AB=c,\angle BAC=A,\angle ABC=B,\angle BCA=C,\angle ADE=\alpha$ we have $$DP=PA\cdot \frac{\sin B}{\sin \alpha}=a\cdot \frac{\sin B}{\sin \alpha},$$$$DE=AE\cdot \frac{\sin A}{\sin \alpha}=\frac{bc}{a+b} \cdot \frac{\sin A}{\sin \alpha}.$$So, $$\frac{DP}{DE}=\frac{a+b}{c}.$$With$\frac{BC}{BE}=\frac{a}{\frac{ac}{a+b}}=\frac{a+b}{c},$ we get $$\frac{DP}{DE}=\frac{BC}{BE}.$$Hence, $$\frac{DP}{BC}=\frac{DE}{BE}=\frac{\sin \angle EBD}{\sin \angle EDB}=\frac{\sin \angle DBC}{\sin \angle BDP},$$which is enough to show that $BD\parallel CP.$

RC. wrote: China owns the license to weird problems. Let \(\odot BDC \cap AB = F\) and \(\odot CDE \cap AB = G\), in \(\angle DFA = \angle C, \angle DGE = \angle DCE = \angle C/2 \Rightarrow \Delta DFG\) is isosceles and \(DF = FG\) but due to cyclic quadrilateral \(\square BCDF, CD = DF\) so, \(CD=DF=FG.\) Now, note that \(\Delta ADP \cong \Delta BGC\) (ASA - congruency axiom). \(\Rightarrow CG = PD \Rightarrow \Delta BFC \cong \Delta APC \Rightarrow \angle BCF = \angle ACP \Rightarrow \measuredangle {[BC,PC]} = \measuredangle {B/2} \quad \square\) What motivates you to consider constructing point $F$ and $G$? It seems like a really arbitrary point one would consider?

$B_0,C_0$ is midpoint of arc $AC,AB$. +)Apply Pascal theorem for $AB_0CBC_0$ =>$ P, B_0, C_0$ are colinear. +) Easy to see $BC=PA=PI$ +) $PI^2=PA^2=PB_0PC_0$ => triangle $PB_0I$ and $PIC_0$ are similar =>$PI \parallel BC$ =>$PIBC$ is parallelogram =>$ Q.E.D$

Let $F, F'$ be the feet of the internal, external angle-bisectors of $\angle BAC$ onto $BC.$ Let $I$ be the incenter of $\triangle ABC$ and $M$ be the midpoint of $FF'.$ It's well-known that $AMF$ is an isosceles triangle. The key claim is that $BIPC$ is a parallelogram. One way to see that this must be true is to let $I'$ be the point so that $BI'PC$ is a parallelogram. Then we have $AP = BC = I'P$, and so $API'$ is an isosceles triangle. Hence, we have that $\triangle API' \sim \triangle AMF \Rightarrow I' \in AF.$ However, $BI' || CP$ means that we should have $I' \in BD$ if the problem is to be believed, which would uniquely determine $I'$ as the incenter $I.$ With the above reasoning, we see that the problem is equivalent to $BIPC$ being a parallelogram, which we'll now show. Lemma 1. $IP || BC.$ Proof. Let $P' \in AM$ such that $IP' || BC.$ Let $X = ED \cap AI.$ We have that $$(E, D; X, P) =^{A} (B, C; F, M) = \frac{\frac{BF}{FC}}{\frac{BM}{MC}} = - \frac{CF}{FB} = (C, B; F, \infty_{BC}) =^{I} (E, D; X, P'),$$where $\infty_{BC}$ denotes the point at infinity along line $BC.$ This clearly implies that $P = P'$, and so the lemma is proven. $\blacksquare$ Due to the lemma, we have from $AA$ Similarity that $\triangle API \sim \triangle AMF$, and so in particular $IP = AP = BC.$ With the lemma, we now conclude that $BIPC$ is a parallelogram, as desired. $\square$

Let $ BD \cap CE=I $ . $\frac{EI}{IC}=\frac{BE}{BC}=\frac{BE}{AP}=\frac{BE \cdot sinB}{DP \cdot sin\angle ADE}=\frac{BE \cdot DE \cdot sinB}{DP \cdot AE \cdot sinA}=\frac{DE}{DP}$ . $\square$