Incenter and midpoints! Boom!

The areas can only be equal if B and Q coincides since A, B, Q are collinear. Similarly C coincides with P.
That means Median PE and QD, that is CE and BD meet at centroid I.
Centroid and incenter coincides.
That is possible only in equilateral triangle.
Angle BAC=60.

First of all, we note the following facts: $\begin{aligned} &1. \quad 2\cdot S_{ABC}= AB\cdot AC \sin \angle BAC\\ &2. \quad 2\cdot S_{APQ}=AP\cdot AQ \sin \angle BAC\\ &3. \quad 0\leq AP\leq AC\\ &4. \quad 0\leq AQ\leq AB\\ \end{aligned}$

Note that $3.$ and $4.$ imply that $AP\cdot AQ \leq AB\cdot AC \quad (5.)$ with equality holding only if it holds in both $3.$ and $4.$ .

The question states that $S_{ABC}=S_{APQ}$ . Hence, equality holds in $5.$ . Hence, equality also holds in $3.$ and $4.$ , that is, $P=C$ and $Q=B$ . Therefore, the incentre and the centroid of $\Delta ABC$ coincide.

This implies that $\Delta ABC$ is equilateral. Hence $\angle BAC$ is $60^{\circ}$ .