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## Romanian TST 2011 Problem 3

Let $ABC$ be a triangle with $AB < AC$. The perpendicular bisector of the side $BC$ intersects $AC$ in $D$. The angle bisector of $\angle ADB$ intersects the circumcircle of $\Delta ABC$ in $E$. Prove that the angle bisector of $\angle AEB$ is perpendicular to the line which connects the incenters of the triangles $\Delta ADE, \ \Delta BDE$.

Romanian TST 2011

Proof: First note that the angle bisector of $\angle ADB$ is parallel to $BC$. Extent $DE$ until it meets the circumcircle of $ABC$ again in $F$. Then we have $\angle BED=\angle EFC=\angle EAD$. This means that the triangles $\Delta ADE,\ \Delta EDB$ are similar. Consider their incenters $X,Y$, respectively. Then it is easy to see that $\Delta XDY$ and $\Delta ADE$ are similar. Moreover, the triangle $\Delta XDY$ is rotated with an equal angle, but different direction with respect to both triangles $\Delta ADE,\ \Delta EDB$. This means that $XY$ forms equal angles with $AE$ and $BE$. From this it follows that the angle bisector of $\angle AEB$ is perpendicular to $XY$, and we are done.