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This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with circle geometry, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it or draw the diagram. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Let $ABC$ be a triangle. The bisector of $\angle BAC$ meets the side $BC$ at point $D$, and meets the circumcircle of the triangle at point $E$. The line through $B$ that is parallel to line $EC$ meets side $AC$ at point $F$. Prove that line $EB$ is tangent to the circumcircle of triangle $ADF$ at point $B$.

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Since $EC||BF$, we see that $$\measuredangle FBD=\measuredangle BCE=\measuredangle BAE=\measuredangle DAF,$$ which says that $ABDF$ is cyclic.

Now, $\measuredangle DBE=\measuredangle EAC=\measuredangle BAD,$ which says that $BE$ is a tangent to the circumcircle of $ABDF$.

Done!

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Hint1. Please include some real effort, you just pasted the intro of your previous question.

Hint2. For drawing diagrams, Geogebra is very efficient.

enter image description here

Hint3. $EB=EC$ since $AE$ is a bisector, hence in the given configuration there are a lot of equal angles. I have marked them in red. To prove the claim by angle chasing is now very simple: you just have to show that $ED\cdot EA=EB^2$, or $\frac{ED}{EB}=\frac{EB}{EA}$.

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