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$I$ is the incenter of $\triangle ABC$. Lines $BI, CI$ meet the line parallel to $BC$ through $A$ at points $D, E$. The perpendicular bisectors of segments $BD, CE$ meet $BC$ at points $X, Y$.

a) Prove that $XI=YI=AI$.

Consider two points on the circumcirlce of $\triangle ABC$, called $P, Q$, such that $\angle DPE= \angle DQE= 90^{\circ}$.

b) Prove that $PQ$ bisects segment $XY$.

Note that by simple angles, triangles $ABD, ACE$ are both isosceles. So the perpendicular bisectors meet at $A$. But how to continue? And also note that $P, Q$ are on a circle with diameter $DE$.

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$ABYD$ is a rhombus, and $ACXE$ is a rhombus too. Since $I$ belongs to the diagonal $BD$, $IA=IY$. Since $I$ belongs to the diagonal $CE$, $IA=IX$, hence $IA=IX=IY$ by transitivity.

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Since $AD=AB$ and $AE=AC$ we know that $DE=b+c$ and we also know the positions of $X,Y$ with respect to $B,C$ in terms of the side lengths. The $PQ$-line is the radical axis of two circles, hence it is orthogonal to the line joining the midpoint $M$ of $DE$ with the circumcenter $O$ of $ABC$. $MP=MQ=\frac{b+c}{2}$ and you have enough informations to prove that $PQ\cap BC$ is the midpoint of $XY$ by considering the powers of the involved points with respect to the circle centered at $M$ and the circle centered at $O$.

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  • $\begingroup$ Thanks! I tried, but it didn't work out. Can you detail your amazing solution for b)? $\endgroup$ – Leo Gardner Apr 12 '18 at 17:54

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