In $\triangle ABC$, $D$ bisects $AB$, $E$ trisects $BC$ and $\angle ADC=\angle BAE$. Find $\angle BAC$

Question

Giving a $\triangle ABC$ having point $D$ as the mid point of $AB$ and $E$ as the trisection point on $BC$ such that $BE>CE$. If $\angle ADC=\angle BAE$. Find $\angle BAC$

I can't solve this problem but I was able to construct the picture of it and it seems that $\angle BAC=90^\circ$. In the picture below is what I could find so far ( there is a tons of parallel but I don't add it to make to picture more comfortable). Any idea?

Answer 1

Let X be the intersection of AE and CD. Evaluate

$$\frac{DX}{XC}= \frac{Area_{ADE}}{Area_{AEC}}=\frac{\frac12\frac23 Area_{ABC}}{\frac13 Area_{ABC}}=1 $$

Then, with $\angle ADC=\angle BAE$, we get $AX= DX=CX$. So, $A$, $D$ and $C$ are cyclic with $X$ being the circumcenter. Thus, $\angle BAC = 90^\circ$.