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.
 A: 
In the above construction, $X'$ is the image of $X$ with respect to a counter-clockwise rotation of $90^\circ$ centered at $A$. We have $\color{green}{AP}=\color{green}{AP'}$ and $AP\perp AP'$ by construction. Since $$\color{purple}{PQ}=BP+DQ=QD+DP'=\color{purple}{QP'},$$ the quadrilateral $APQP'$ is a kite and $PP'\perp AQ$. The claim ($\widehat{QAP}=45^\circ$) easily follows.
A: Using the cosine rule:$$|AP|^2+|AQ|^2-2|AP||AQ|cos\alpha=|PQ|^2$$Replacing $|AP|,|AQ|$ and $|PQ|$:$$|AP|=\sqrt{a^2+r^2}$$$$|PQ|=\sqrt{b^2+r^2}$$$$|PQ|=a+b$$
Here is $a=|BP|, b=|DQ|$:$$a^2+r^2+b^2+r^2-2\sqrt{a^2+r^2}\sqrt{b^2+r^2}cos\alpha=(a+b)^2$$$$\implies2r^2-2\sqrt{a^2+r^2}\sqrt{b^2+r^2}cos\alpha=2ab$$Note that $b =\frac{r^2-ar}{r+a}$. Substituting $b$ and simplifying further:$$cos\alpha=\frac{r^2(r+a)-a(r^2-ar)}{\sqrt{a^2+r^2}\sqrt{2r^2(a^2+r^2)}}\implies cos\alpha=\frac{1}{\sqrt{2}}$$$$\implies \alpha=45°$$
A: 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 $T$. 
A: Let $R$ be on $PQ$ such that $\overline{RP} = \overline{BP}$ and $\overline{RQ} = \overline{DQ}$ (such a point can be chosen because $\overline{PQ} = \overline{BP}+\overline{DQ}$). Let $l$ be the line perpendicular to $PQ$ at $R$, and let $A'$ be on $l$ such that $A'$ and $A$ are on the same side of $PQ$, and $\overline{A'R} = \overline{AB} = \overline{AD}$.
Since $\overline{A'R} = \overline{AB}$, $\angle A'RP = 90^{\circ} = \angle ABP$, and $\overline{RP} = \overline{BP}$, by SAS we have $\triangle A'RP\cong \triangle ABP$, and hence $\overline{A'P} = \overline{AP}$. By a similar argument, we have $\triangle A'RQ\cong \triangle ADQ$, and hence $\overline{A'Q} = \overline{AQ}$. BY SSS, we have $\triangle A'PQ\cong \triangle APQ$, and since $A'$ and $A$ lie on the same side of $PQ$, this forces $A' = A$.
Thus, we have $\triangle ARP\cong \triangle ABP$ and $\triangle AQP\cong \triangle ADP$, so $\angle RAP = \angle BAP$ and $\angle RAQ = \angle DAQ$. We thus have
$$90^{\circ} = \angle DAB = \angle DAQ + \angle RAQ + \angle RAP + \angle BAP = 2(\angle RAQ + \angle RAP) = 2\angle QAP$$
from which the claim follows.
