Proving $\angle QAP=45^\circ$ if $ABCD$ is a square with points $P$ in $BC$, $Q$ in $CD$ satisfying $\overline{BP}+\overline{DQ}=\overline{PQ}$ Here is the problem:

Let $ABCD$ be a square with points $P$ in $BC$, $Q$ in $CD$ satisfying $\overline{BP}+\overline{DQ}=\overline{PQ}$. Prove that $\angle QAP=45^\circ$.

So far I have been trying to show that $\overline{BP}=\overline{DQ}$ so that the sum of angles on both sides of $\angle QAP$ is $45^\circ$
Any hint or guidance will be great, Thanks in advance.

Perform a $90^{\circ}$ rotation of the square around the vertex $A$ in a counterclockwise direction. Then the image $P_1$ of $P$ lies on the extension of edge $CD$ so that $P_1D = PB$ and $D$ is between $C$ and $P_1$. Now triangles $ABP$ and $ADP_1$ are images of each other under the rotation so they are congruent and $AP$ is orthogonal to $AP_1$. Moreover $AP = AP_1$. Now, $$P_1Q = P_1D + DQ = PQ + DQ = PQ$$ Therefore triangles $P_1QA$ and $PQA$ are congruent and $\angle \, PAQ = \angle \, P_1AQ$. As $\angle \, PAP_1 = 90^{\circ} = \angle \, PAQ + \angle \, P_1AQ = 2 \, \angle \, PAQ$, this means that $\angle \, PAQ = 45^{\circ}$.  
Or, if you take the longer way, which reveals some more properties, $\angle \, AQP_1 = \angle \, AQD = \angle \, AQP = \alpha.$ 
Let $T$ be the point on $PQ$ such that $BP = TP$, then  Then $DQ = TQ$. Since $\angle \, AQD = \angle \, AQP = \angle \, AQT = \alpha$, the triangles $AQD$ and $AQT$ are congruent meaning that $\angle \, ADQ = \angle \, ATQ = 90^{\circ}$. Which combined with the fact that $PT = PB$ implies that triangles $ABP$ and $ATP$ are congruent. So we conclude that  $\angle \, BAP = \angle \, TAP$. Combined with $\angle \, DAQ = \angle \, TAQ$ this means that $\angle \, PAQ = \frac{1}{2} \, \angle \, BAD = 45^{\circ}$.
Observe that $AT = AD = AB$ and $AT$ is orthogonal to $PQ$. Thus the circle centered at $A$ and of radius $AB$ is tangent to $PQ$ at $T$. 
