Solution:
First we will show that points $P$ and $Q$ are not on the line segment $AB$.
Assume that $Q$ is on the line segment $AB$. Since $CI = QI$ and $\angle IBQ = \angle IBC$, either the triangles $CBI$ and $QBI$ are congruent or $\angle ICB + \angle IQB = 180^\circ$. In the first case, we have $BC = BQ$ which contradicts $AB$ being the shortest side.
In the second case, we have $\angle IQA = \angle ICB = \angle ICA$ and the triangles $IAC$ and $IAQ$ are congruent. Hence this time we have $AC = AQ$, contradicting $AB$ being the shortest side.

Case 1

Now we will show that the lines $PE$ and $CQ$ are perpendicular.
Since $\angle IBQ \leq \angle IAB = (\angle CAB)/2 < 90^\circ$ and $\angle ICB = (\angle ACB)/2 < 90^\circ$, the triangles $CBI$ and $QBI$ are congruent. Hence $BC = BQ$ and $\angle CQP = \angle CQB = 90^\circ - (\angle ABC)/2$. Similarly, we have $AC = AP$ and hence $BP = AC - AB$.
On the other hand, as $DE = CD$ and $CD + AC = u$, where $u$ denotes the semiperimeter of the triangle $ABC$, we have $BE = BC - 2(u - AC) = AC - AB$. Therefore $BP = BE$ and $\angle QPE = (\angle ABC)/2$.
Hence, $\angle CQP + \angle QPE = 90^\circ$.