Without restriction, let $|A C|>|A B|$. Since $D$ and $E$ are the midpoints of the sides $\overline{A B}$ and $\overline{A C}$, by the intercept theorem, $D E$ is parallel to $B C$. In the case $E=P$ or $D=Q$, $C E D B$ would be a cyclic quadrilateral with parallel sides $B C$ and $D E$, thus an isosceles trapezoid with $|B D|=|E C|$, and therefore $|A B|=|A C|$, which is excluded. If it can be shown that the (non-degenerate) triangles $A P C$ and $B Q A$ are similar, the claim follows, since due to $\angle P Q A=\measuredangle P D A=180^{\circ}-\measuredangle B D P=\measuredangle P C B=\measuredangle P C A+\gamma=\measuredangle B A Q+\measuredangle A E D=\measuredangle D P Q+\measuredangle A P D=$ $\measuredangle A P Q$ (using the inscribed angle theorem), the triangle $A P Q$ is isosceles with $|A P|=|A Q|$. - Proof that triangles $A P C$ and $B Q A$ are similar: First, by the inscribed angle theorem,

$$\measuredangle D Q B=360^{\circ}-\measuredangle E Q D-\measuredangle B Q E=\measuredangle D A E+\measuredangle E C B=\alpha+\gamma=180^{\circ}-\beta=180^{\circ}-\measuredangle E D A=\measuredangle A P E,$$

$\measuredangle E P C=\measuredangle D P C-\measuredangle D P E=\left(180^{\circ}-\beta\right)-\alpha=\gamma=\measuredangle A E D=\measuredangle A Q D$.
From this, the similarity of triangles $A P C$ and $B Q A$ follows:

1. Proof: Let $Q^{\prime}$ be the uniquely determined point on the same side of $A C$ as $P$, such that triangle $A Q^{\prime} C$ is similar to triangle $B Q A$. Since $D$ and $E$ are the midpoints of the sides $\overline{A B}$ and $\overline{A C}$, the sub-triangles $A Q^{\prime} E$ and $B Q D$ as well as $C E Q^{\prime}$ and $A D Q$ are also similar. Therefore, $Q^{\prime}$ lies on the circumcircles of $A E P$ and $C E P$, which intersect at $E$ and $P$. $Q^{\prime}=E$ is excluded because $Q \neq D$. Thus, $Q^{\prime}=P$, and the triangles $A P C$ and $B Q A$ are similar. 2. Proof: By the sine rule in $A D Q$ and $B Q D$, $|A Q|:|A D|=\sin \measuredangle Q D A: \sin \measuredangle A Q D,|B Q|:|B D|=$ $\sin \measuredangle B D Q: \sin \measuredangle D Q B$. With $\measuredangle B D Q=180^{\circ}-\measuredangle Q D A$, it follows that $|A Q|:|B Q|=\sin \measuredangle D Q B: \sin \measuredangle A Q D$. Similarly, $|C P|:|P A|=\sin \measuredangle A P E: \sin \measuredangle E P C$. Due to (1) and (2), $A P C$ and $B Q A$ are similar by the criterion sws.