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In $\triangle ABC$, perpendiculars from $A$ to the bisectors of angle $B$ and angle $C$ meet the bisectors in $D$ and $E$, respectively. The line through $D$ and $E$ intersects $AC$ at $X$ and $AB$ at $Y$. What fractional part of the area of $\triangle ABC$ is the area of region $XYBC$?

What I've tried: I kind of have no idea how to approach this problem. At first I thought $AD$ might equal $AE$ since $AD = c \cdot \sin {ABD}$, and $AE = b \cdot \sin {ACE}$, but if you use Law of Sines it actually says $c \cdot \sin {(2 \cdot ABD)} = b \cdot \sin {(2 \cdot ACE)}$ (since $ABC = 2 \cdot ABD$ and $ACB = 2 \cdot ACE$). If you work that out, it actually implies that $AD \neq AE$ unless the triangle is isosceles.

I've had no more good ideas. Can anyone help?

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Let $BD\cap CE=\{O\}$.

Thus, $AEOD$ is cyclic, which says that $$\measuredangle ADX=180^{\circ}-\measuredangle ADE=180^{\circ}-\measuredangle AOE=180^{\circ}-\left(\frac{\alpha}{2}+\frac{\gamma}{2}\right)=90^{\circ}+\frac{\beta}{2}.$$ Thus, $\measuredangle MDX=\frac{\beta}{2}$, which says that $XY||BC$.

Now, by law of sines for$\Delta ABC$ and for $\Delta ADX$ we obtain: $$\frac{AX}{b}=\frac{AX}{\frac{c\sin\beta}{\sin\gamma}}=\frac{AX}{\frac{AD\sin\beta}{\sin\frac{\beta}{2}\sin\gamma}}=$$ $$=\frac{\sin\frac{\beta}{2}\sin\gamma}{\sin\beta}\cdot\frac{\sin\left(90^{\circ}+\frac{\beta}{2}\right)}{\sin\gamma}=\frac{1}{2}.$$ Thus, $$S_{\Delta AXY}=\frac{1}{4}S_{\Delta ABC}$$ and $$S_{XYBC}=\frac{3}{4}S_{\Delta ABC}.$$ Done!

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