Which excircles are being referred to?

AshAuktober wrote: Which excircles are being referred to? referred to each side (It is unique since there are only 3 of them)

Let $D$, $E$, $F$ be points there excircles touch sides $BC$, $CA$, $AB$ and $M$ be midpoint of segment $XY$. Also let $U,V,R$ be projections of points $X,Y,M$ on $AC$. We need to prove that $RA=RC$. Note that $UR=VR$, so it's enough to prove that $AV=CU$. Let $\alpha=\angle BAC$, $\gamma = \angle ACB$, $a=BC$, $b=CA$, $c=AB$, $p=(a+b+c)/2$. Then $AV=AY \cdot cos (\alpha /2)$ and $CU=CX\cdot cos (\gamma/2)$ so $AV=CU \iff AY \cdot cos (\alpha /2) = CX\cdot cos (\gamma/2)$. Let's work with $LHS$. Define $K= DE \cap AB$ (points in order $A-B-K$ and $E-D-K$). By Menelaus's theorem it is not hard to find that $AK=\frac{c(p-b)}{a-b}$. Note that $AY$ is bisector of $\angle KAE$. So (by known formula) \[ AY = \frac{2 \cdot AE \cdot AK \cdot cos(\alpha/2)}{AE+AK}=\frac{c(p-b)(p-c) \cdot 2cos(\alpha/2)}{p(a+c-b)-ac}\]and thus \[AY \cdot cos(\alpha/2)=\frac{c(p-b)(p-c) \cdot 2cos^2(\alpha/2)}{p(a+c-b)-ac}=\frac{c(p-b)(p-c) \cdot (cos (\alpha)+1)}{p(a+c-b)-ac}\]which is by law of cosines equals to \[\frac{(p-b)(p-c)(a+b+c)(b+c-a)}{2b(p(a+c-b)-ac)}=\frac{4p(p-a)(p-b)(p-c)}{2b(p(a+c-b)-ac)}\]what is symmetric onto $a$ and $c$, so $CX \cdot cos (\gamma/2)$ equals to same expression and we are solved the problem.

Let $A_1$, $B_1$, and $C_1$ be the respective extouch points. Let $CI$ meet $A_1B_1$ at $P$ and let $AI$ meet $C_1B_1$ at $Q$. Let $P'$ and $Q'$ be the feet of $P$ and $Q$ onto $AC$. Claim: The midpoint of $PQ$ is equidistant from $A$ and $C$ It is sufficient to show that $CP'=AQ'$. We compute $$CP'=\cos(\tfrac{C}{2})\cdot CP=\cos^2(\tfrac{C}{2})\cdot \frac{2 CA_1\cdot CB_1}{CA_1+CB_1}=\frac{1+\cos(C)}{2}\cdot \frac{(a-b+c)(-a+b+c)}{2c}=$$$$\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{8abc}$$As this expression is symmetric the claim follows. It is a well known fact that the midpoint of $IB_1$ is also equidistant from $A$ and $C$. Thus the Newton-Gauss line of complete quadrilateral $B_1QXYIP$ is the perpendicular bisector of $AC$, and the result follows (I guess I should also add that $IB_1$ and $PQ$ do not share a midpoint as it would cause $X$ and $Y$ not to exist).

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Let $A$-excircle meet $AB,AC$ at $E,D$. Let $C$-excircle meet $BC,AC$ at $G,F$. Let $K,L$ be projections of $X,Y$ on $AC$. Claim $: KF=DL$. Proof $:$ Note that since $\angle KYF = \angle I_CFG = \frac{C}{2}$ we have $KF = YF.\sin{\frac{C}{2}}$ and similarly we have $DL = XD.\sin{\frac{A}{2}}$ so we need to prove $\frac{YF}{XD} = \frac{\sin{\frac{A}{2}}}{\sin{\frac{C}{2}}}$. Note that $\angle FYA = \angle KYA - \angle KYF = \frac{B}{2}$ and similarly $\angle DXC = \frac{B}{2}$. Note that $YF = FA.\frac{\sin{\frac{A}{2}}}{\sin{\frac{B}{2}}}$ and $XD = DC.\frac{\sin{\frac{C}{2}}}{\sin{\frac{B}{2}}}$ and $AF = p-b = CD$ so $\frac{YF}{XD} = \frac{\sin{\frac{A}{2}}}{\sin{\frac{C}{2}}}$ as wanted. Now note that $AK = CL$ so $AY^2 - CY^2 = CX^2 - AX^2$ so $AY^2 + AX^2 = CX^2 + CY^2$ so if $P$ is the midpoint of $XY$ then $AP = CP$ as wanted.

Funny. Notice that by newton gauss line it suffices to prove that if $AI \cap B’C’=P$ and $CI \cap A’B’=Q$ then midpoint $PQ$ lies on perpendicular bisector of $AC$, let the perpendiculars from $P,Q$ to $AC$ be $U,V$ then as $OM \perp AC$ so it suffices to show $P$ and $Q$ has same power WRT $(ABC)$; let $f(\cdot)=Pow_{(AB_1C_1)}-Pow_{(ABC)}$ which is linear as well known; now linearity of PoP yields that $f(P)$ is symmetric; so we are done.