Since triangle $APQ$ is isosceles, we have $\angle APQ = \angle AQP$. $BMIP$ and $CMIQ$ are cyclic quadrilaterals, so $\angle APQ = \angle BMI$ and $\angle AQP = \angle CMI$. It follows that $\angle BMI = \angle CMI$. (1).

Let $K$ be the point where lines $IM$ and $BC$ meet. Since $BI$ is the bisector of $\angle ABC$ and $BMIP$ is a cyclic quadrilateral, we get $\angle IMP = \angle IBP = \angle IBK$. From $\angle KMD = \angle KBD$ we infer that $BDKM$ is cyclic, so $\angle IDK = \angle BMK$. Similarly, we prove that $CEKM$ is also cyclic and $\angle IEK = \angle CMK$.

Using (1), it follows that $\angle IDK = \angle IEK$.
The quadrilaterals $BDKM$ and $BMIP$ are cyclic, so $\angle BKD = \angle BMD = \angle BIP$, (2).
Also, the quadrilaterals $CEKM$ and $CMIQ$ are cyclic, so $\angle CKE = \angle CME = \angle CIQ$, (3).
From (2) and (3) we deduce that $\angle BKD + \angle CKE = \angle BIP + \angle CIQ$. Consequently, we obtain $\angle DKE = 180^\circ - (\angle BKD + \angle CKE) = 180^\circ - (\angle BIP + \angle CIQ) = \angle DIE$. Since $\angle IDK = \angle IEK$ and $\angle DKE = \angle DIE$, it follows that $DIEK$ is a parallelogram, so line $KI$ bisects the segment $DE$. As points $I$, $K$, and $M$ are collinear, the conclusion follows.