parmenides51 wrote: In the triangle $ABC$, $D$ is the midpoint of $AB$, and $E$ is the point on the side $BC$, for which $CE = \frac13 BC$. It is known that $\angle ABC =\angle BAE$. Find $\angle BAC$. Why the data of point D, if that point is not used further?

indeed I had a typo, the condition is $\angle ADC =\angle BAE$ and not $\angle ABC =\angle BAE$ thanks for noticing

Given $\triangle ABC\ :\ A(-a,0),B(b,0),C(0,c)$. Midpoint $D(\frac{b-a}{2},0)$, point $E(\frac{b}{3},\frac{2c}{3})$. Condition gives $a=0$, so $\angle BAC=90^{\circ}$.