Let $I$ be the incenter. Then $\angle IDE = \angle IAE$ from $ADIE$ concyclic. Then $\angle IAE = \angle CAE-\angle CAI = 90-\dfrac{1}{2}\angle C - \dfrac{1}{2}\angle A = \dfrac{1}{2}\angle B$ so $\angle IDE = \angle IBC$, so $DE\parallel BC$.

If $AD,AE$ hits $BC$ at $K,L$, then $K,L$ is the reflection of $A$ on the $B$-bisector and $C$-bisector, resp., and consequently $DE$ is the image of $KL$ after a homothety from $A$ with ratio $1/2$, implying $DE||KL\equiv BC$.

Dear Mathlinkers, have a look at http://jl.ayme.pagesperso-orange.fr/Docs/An%20unlikely%20concurrence.pdf p. 3 Sincerely Jean-Louis

We're done by the right angles on intouch chord (well, midline in this case) lemma.

Let $X,Y$ be feet of $A$ on angle bisectors of $B$ and $C$ respectively. We have \[\angle IXY = \angle IAY = \angle CAY - \angle CAI = 90^{\circ} - \frac{C}{2} - \frac{A}{2} = \frac{B}{2}.\]

It's enough to prove $ \angle EDB $ = $ \angle DBC $ . Claim 1 : AEID is cyclic . Proof : $ \angle AEC $ + $ \angle ADB $ = $ 180^\circ $ . Hence, $ \angle EDI $ = $ \angle IAE $ = $ \angle BAI $ - $ \angle BAE $ . $ \angle BAE $ = $ 90^\circ $ - $ \angle AFC $ and $ \angle AFC $ = $ 180^\circ $ - $ \angle BAC $ - $ \angle FCA $ . (from the sum of the angles of the triangle FAC) We know that $ \angle BAC $ + $ \angle ABC $ +$ \angle ACB $ = $ 180^\circ $ (*) And $ \angle ABC $ = 2$ \angle ABG $ = 2$ \angle GBC $ ,...(**) Due to the (*) and (**) , the proof is complete !

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