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Tangent circles
baopbc   1
N Dec 25, 2016 by baopbc
Source: Own - Proposes for APMC (#1)
Given triangle $ABC$ with the inner - bisector $AD$. The line passes through $D$ and perpendicular to $BC$ intersects the outer - bisector of $\angle BAC$ at $I$. Circle $(I,ID)$ intersects $CA$, $AB$ at $E$, $F$, reps. The symmedian line of $\triangle AEF$ intersects the circle $(AEF)$ at $X$. Prove that the circles $(BXC)$ and $(AEF)$ are tangent.
Diagram
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baopbc
Dec 25, 2016
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Dec 25, 2016
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Source: Own - Proposes for APMC (#1)
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baopbc
#1 h  Dec 25, 2016, 12:46 AM • 7 Y
Y by aopser123, mihajlon, buratinogigle, worstsol, sholly, tiendung2006, Adventure10
Given triangle $ABC$ with the inner - bisector $AD$. The line passes through $D$ and perpendicular to $BC$ intersects the outer - bisector of $\angle BAC$ at $I$. Circle $(I,ID)$ intersects $CA$, $AB$ at $E$, $F$, reps. The symmedian line of $\triangle AEF$ intersects the circle $(AEF)$ at $X$. Prove that the circles $(BXC)$ and $(AEF)$ are tangent.
Diagram
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#3 h  Dec 25, 2016, 1:30 AM • 7 Y
Y by buratinogigle, aopser123, mihajlon, CaptainCuong, sholly, Adventure10, Mango247
Solution :

Since $I$ lies on the perpendicular bisector of $EF$ and the outer - bisectore of $\angle AEF$ so $A$, $I$, $E$, $F$ are concyclic.
Hence, $AI$ is the radial axis of $\odot (ID)$ and $\odot (AEF)$.
By radial axis $AI$, $EF$, $BC$ are concurrent at $S$.

Since $S$ is the bisector - foot of $\angle EAF$ so $\tfrac{SF}{SE}=\tfrac{AF}{AE}$
Let $T\equiv XS\cap \odot (AEF)$.
Since $AX$ is the symedian so $\tfrac{XF}{XE}=\tfrac{AF}{AE}=\tfrac{SF}{SE}$.
Hence, $XT$ is the outer - bisector of $\angle ETF$ $\implies A$, $D$, $T$ are collinear.

On ther other hand, $SD$ tangents to $\odot (EDF)$ so $SD^2=\overline{SF}\cdot \overline{SE}=\overline{SX}\cdot \overline{ST}$
$\implies SD$ tangents to $\odot (DTX)$.
$\implies \angle XDB=\angle DTX=\angle XFB$ $\implies B$, $F$, $D$, $X$ are concyclic.
Similarly, $C$, $D$, $E$, $X$ are concyclic.

So we get the angle chasing, $\angle BXF=\angle BDF=\angle FED=\angle FEX-\angle DEX=\angle FEX-\angle BCX$.
Hence, $\odot (BXC)$ and $\odot (AEF)$ are tangent. $\square $
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