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I have the following diagram. I am trying to prove that $\angle BAE=\angle DAC$. I've tried to show this via angle chasing and similar triangles, but I am stuck. I am hoping that someone can at least give a direction to head in. Diagram

So far I've been able to show the following:

$\triangle ABE\sim\triangle AGB\sim\triangle BGE$

$\triangle AHC\sim\triangle CHE\sim\triangle ACE$

$\triangle ABC\sim\triangle DGH$

Also I suspect these things are true, but I have been unable to prove them.

$\triangle AFB\sim\triangle CFD$

$\triangle AFC\sim\triangle BFD$

$\triangle ABG\sim\triangle ACD$

$\triangle ADH\sim\triangle AGD$

A circle with diameter $AD$ passes through $B$ and $C$

Thank you!

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(As you've noticed) Since the pairs of opposite angles of $ABCD$ sum up to $\pi$, $ABCD$ is a cyclic quadrilateral. Hence, $\angle CAD=\angle CBD$. But, as you've already noticed, $\angle CBD=\angle CBE=\angle BAE$.

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    Ah thanks. I actually didn't know the property of cyclic quadrilaterals that gives you CAD=CBD. But a quick wikipedia search helped fill in that gap. – 2017-02-06
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    It's just the fact that they are angles at the circumference with the same arc. I don't know if it has a special formulation in this case. – 2017-02-06