In triangle $\triangle ABC$, $BC>AC$, $I_B$ is the $B$-excenter, the line through $C$ parallel to $AB$ meets $BI_B$ at $F$. $M$ is the midpoint of $AI_B$ and the $A$-excircle touches side $AB$ at $D$. Point $E$ satisfies $\angle BAC=\angle BDE, DE=BC$, and lies on the same side as $C$ of $AB$. Let $EC$ intersect $AB,FM$ at $P,Q$ respectively. Prove that $P,A,M,Q$ are concyclic.
Problem
Source: Bulgarian TST 2020 P6
Tags: geometry
Steve12345
24.01.2021 01:34
Let $S$ be the intersection of the bisectors of the segments $CE$ and $BD$ (the center of rotation). Notice the cyclic quads $CBSP$ and $EDSP$ by spiral similarity. From these, let: $x=\angle BSC=\angle BPC=\angle PCF$. Notice that for $PAMQ$ to be cyclic we need $\angle QMI_B = x =\angle FMI_B =\angle CSB$. From simple angles we have $\angle FI_BM = \angle CBS$ so I was motivated to prove $\triangle FI_BM \sim \triangle CBS$ (since this must be true if the problem is true). So we need: $\frac{FI_B}{I_BM}=\frac{a}{BS}$ $(\star)$. I saw it was bashable, so I bashed: $s=\frac{a+b+c}{2}$, $\triangle = \sqrt{s(s-a)(s-b)(s-c)}$ - Area of $\triangle ABC$. $FI_B$: By LoS on $\triangle FI_BC$ we get: $\frac{FI_B}{\sin(\frac{\alpha-\beta}{2})}=\frac{a}{\sin{\frac{\alpha}{2}}}$ so $\boxed{ FI_B=\frac{a \cdot \sin(\frac{\alpha-\beta}{2})}{\sin{\frac{\alpha}{2}}} }$ ($FC=CB=a$) $I_BM$: Looking at triangle $AI_BC$ we get $I_BM =\frac{AI_B}{2}=\frac{r_B}{2\cdot \cos(\frac{\alpha}{2})}=\boxed{ \frac{\triangle}{2\cdot (s-b)\cdot \cos(\frac{\alpha}{2})} } $ $BS$: Looking at triangle $BSD$ we get: $\boxed{ BS=\frac{s-c}{2\cdot \sin{\frac{\alpha-\beta}{2}}} }$ Plugging these into $(\star)$ the equation is satisfied, and we're done.