contest-matheuclidean-geometrygeometrytriangles
I recently found a lemma that
$\textbf{Lemma:}$ the line connecting the circumcenter and A-excenter is perpendicular to the line formed by intersections of the B-angle bisector and C-angle bisector with their respective opposite sides.
I tried much but couldn't find a proof.
Can anyone please help me with the proof?
in $ \triangle ABC$, $\angle B=90^\circ$, $AD=DC$
Define $K$ as midpoint of $BC$. Call the intersection of $BS$ and $DF$ as $P$. Since $S$ is the center of the circle of $ABC$, we have $\angle BSC=2\angle A$. Since $SB=SC$, SK bisects $\angle BDF$ therefore $\angle BSK=\angle A$. On the other hand since $D$ and $F$ are feet of the altitudes $CAFD$ is cyclic, hence $\angle BDF = \angle A$. Now $\angle A = \angle BSK = \angle BDF$ tells us $KSPD$ is a cyclic quadrilateral. Hence $\angle SKB=\angle DPS=90^{\circ}$.
Best Answer
Let $P$ and $Q$ be the intersections of the radical axis of the circumcircle and the $A$-excircle with $AB$ and $AC$, respectively. We have $$PB\cdot PA=\operatorname{Pow}_{(O)}(P)=\operatorname{Pow}_{(O_a)}(P)=PF^2,$$ where $F$ is the tangency point of the $A$-excircle to $AB$. Thus, letting $a=BC$, etc. and $s=\frac{AB+BC+CA}{2}=AF$, $$PA(PA-c)=(s-PA)^2\implies PA=\frac{s^2}{2s-c}=\frac{s^2}{a+b}.$$ Similarly $QA=\frac{s^2}{a+c}$. However, we have by the angle bisector theorem that $$AB_b=\frac{bc}{a+c},\ AC_b=\frac{bc}{a+b},$$ so $\triangle AB_bC_b\sim\triangle AQP$. This means $B_bC_b || PQ$, but $PQ\perp OO_a$, so we are done.