# Bisecting geo problem - from Art of Problem Solving

$I$ 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 $DE$. $$ABYD$$ 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. 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 $$O$$.