Given $\triangle ABC\ :\ A(-a,0),B(0,b),C(a,0)$. The line $AP\ :\ y=x+a$ intersects the line $BC\ :\ y=-\frac{b}{a}(x-a)$ in the point $P(\frac{a(b-a)}{a+b},\frac{2ab}{a+b})$. The perpendicular bisector of the segment $AP\ :\ y-\frac{ab}{a+b}=-(x+\frac{a^{2}}{a+b})$ intersects the line $AB\ :\ y=\frac{b}{a}(x+a)$ in the point $Q(-\frac{a(a^{2}+b^{2})}{(a+b)^{2}},\frac{2ab^{2}}{(a+b)^{2}})$. The slope of the line $PQ\ :\ m_{PQ}=\frac{a}{b}$, so $PQ \bot BC$.

Trivial with elementary geometry techniques and angle-chasing?? Call $\angle BAC = \angle BCA = \theta$, thus $\angle ABC = 180 - 2\theta$. We call the intersection of the perpendicular bisector of $AP$ and side $AC$ point $D$, and the intersection of line $QD$ and $AP$ point $E$. First we prove $\triangle QEA \cong \triangle QEP$. It is trivial to see $AE \cong EP$ and $\angle AEQ \cong \angle PEQ$ from problem statement, and $EQ \cong EQ$ by reflective so we proved $\triangle QEA \cong \triangle QEP$ by SAS. Since $\angle PAC = 45$, we know $\angle BAP = \theta - 45$. Now since $\angle AEQ = 90$, we know $\angle AQE = 180 - (\angle QAE + \angle QEA) = 180 - (\theta - 45 + 90) = 135 - \theta$. Since $\angle AQE = \angle PQE$ due to our proof from above, we get $\angle AQP = 2(\angle AQE) = 270 - 2\theta$. We know $\angle BQP = 180 - \angle AQP = 2\theta - 90$. Finally we know $\angle BPQ = 180 - (\angle QBP + \angle PQB) = 180 - (180 - 2\theta + 2\theta - 90) = 90$, which means $QP \perp BC$ as desired.

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Denote $\angle QPA=\angle QAP=\alpha$. Then $\angle ACB=45+\alpha$. Hence we get $\angle QPC=\alpha +\angle APC=\alpha+(180-\angle PAC-\angle PCA)=\alpha+90-\alpha=90$ as desired.