angle bisector and concurrency problem.

Question

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In this figure, line BE is angle bisector of ∠ABC and some point X is on BE. If ∠AEB=∠FDB, then prove ∠EDF is right angle.

Answer 1

Let me try to guess what the correct statement of the problem should be.

Problem: Let the angle bisector of $\angle ABC$ meet $AC$ at $E$, and let $X$ be a point on $BE$. Extend $AX,CX$ to intersect $BC,AB$ at $D,F$ respectively. If $\angle AEB=\angle FDB$, prove that $\angle EDF$ is a right angle.

Extend $EA$ and $DF$ to meet at $Y$. Then $\angle YEB=\angle AEB=\angle FDB=\angle YDB$, so $Y,E,D,B$ concyclic. We now have everything in the diagram (except the point $O$, which is a distraction).

By Ceva's theorem, we have $$ \frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA}=1 $$ since $AD,BE,CF$ are concurrent. On the other hand, by Menelau's theorem, since $Y,F,D$ are collinear, $$ \frac{AF}{FB}\frac{BD}{DC}\frac{CY}{YA}=-1. $$ So $$ \frac{CY}{YA}=-\frac{CE}{EA}. $$ But $BE$ is the angle bisector of $\angle ABC$! The angle bisector theorem gives $$ \frac{CE}{EA}=\frac{BC}{BA} $$ i.e., $Y$ divides $CA$ externally in the ratio $$ CY:YA=BC:BA $$ so $BY$ is the external angle bisector of $\triangle ABC$ at $B$ (the external angle bisector theorem). Thus $\angle EDF=\angle EDY=\angle EBY$ is a right angle (internal and external angle bisectors of $\angle B$ are orthogonal).