Home > Geometry > Romanian TST 2011 Problem 3

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.

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