1. Assign Coordinates:
Let $D$ be on $\overline{AC}$ and let $E$ be on $\overline{BC}$ with $\overline{BC}$ being a horizontal line parallel to the x-axis. We proceed with coordinate bashing. Let point $C$ be the origin. We then let $B=(a,0)$, $A=(0,a)$, $E=(b,0)$, and $D=(0,b)$.

2. Find Equations of Lines:
- The line $\overline{AB}$ can be described by the equation:
$$y = -x + a$$
- The line $\overline{AE}$ can be described by the equation:
$$y = \frac{-a}{b}x + a$$

3. Find Slopes of Perpendicular Lines:
Since $\overline{CP}$ and $\overline{DQ}$ are perpendicular to $\overline{AE}$, they both have a slope of $\frac{b}{a}$. We then use points $D$ and $C$ to find the equations of these two lines:
- The line $\overline{PC}$:
$$y = \frac{b}{a}x$$
- The line $\overline{DQ}$:
$$y = \frac{b}{a}x + b$$

4. Find Points of Intersection:
- The point of intersection of $\overline{AB}$ and $\overline{PC}$ (which is point $P$):
$$\begin{aligned} -x + a &= \frac{b}{a}x \\ a &= x\left(1 + \frac{b}{a}\right) \\ a &= x\left(\frac{a + b}{a}\right) \\ x &= \frac{a^2}{a + b} \end{aligned}$$
Substituting $x$ back into $y = \frac{b}{a}x$:
$$y = \frac{b}{a} \cdot \frac{a^2}{a + b} = \frac{ab}{a + b}$$
Thus, $P = \left(\frac{a^2}{a + b}, \frac{ab}{a + b}\right)$.

- The point of intersection of $\overline{AB}$ and $\overline{DQ}$ (which is point $Q$):
$$\begin{aligned} -x + a &= \frac{b}{a}x + b \\ a - b &= x\left(\frac{a + b}{a}\right) \\ a - b &= x\left(\frac{a + b}{a}\right) \\ x &= \frac{a^2 - ab}{a + b} \end{aligned}$$
Substituting $x$ back into $y = \frac{b}{a}x + b$:
$$y = \frac{b}{a} \cdot \frac{a^2 - ab}{a + b} + b = \frac{ab - b^2 + ab}{a + b} = \frac{2ab - b^2}{a + b}$$
Thus, $Q = \left(\frac{a^2 - ab}{a + b}, \frac{2ab}{a + b}\right)$.

5. Calculate Distances:
- Distance $BP$:
$$\begin{aligned} BP &= \sqrt{\left(a - \frac{a^2}{a + b}\right)^2 + \left(0 - \frac{ab}{a + b}\right)^2} \\ &= \sqrt{\left(\frac{a(a + b) - a^2}{a + b}\right)^2 + \left(\frac{-ab}{a + b}\right)^2} \\ &= \sqrt{\left(\frac{ab}{a + b}\right)^2 + \left(\frac{ab}{a + b}\right)^2} \\ &= \sqrt{2\left(\frac{ab}{a + b}\right)^2} \\ &= \frac{ab\sqrt{2}}{a + b} \end{aligned}$$

- Distance $PQ$:
$$\begin{aligned} PQ &= \sqrt{\left(\frac{a^2}{a + b} - \frac{a^2 - ab}{a + b}\right)^2 + \left(\frac{ab}{a + b} - \frac{2ab}{a + b}\right)^2} \\ &= \sqrt{\left(\frac{a^2 - (a^2 - ab)}{a + b}\right)^2 + \left(\frac{ab - 2ab}{a + b}\right)^2} \\ &= \sqrt{\left(\frac{ab}{a + b}\right)^2 + \left(\frac{-ab}{a + b}\right)^2} \\ &= \sqrt{2\left(\frac{ab}{a + b}\right)^2} \\ &= \frac{ab\sqrt{2}}{a + b} \end{aligned}$$

Since $BP = PQ$, our proof is complete. $\blacksquare$

The final answer is $BP = PQ$