Using similar and congruent triangles, and some circle theorems solves the problem

This is like a 5 min bary bash. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ and remark that $P=(a^2:-b^2+bc:-c^2+bc)$ where we computed the coordinates of $P=\frac{I_B+I_C}2$ using the midpoint formula on normalized $I_B,I_C$. Since $Q$ is on cevian $BP$, it can be parameterized by $Q=(a^2:t:-c^2+bc)$. However, $A,Q,I$ are colinear so $$\begin{vmatrix}1&0&0\\ a^2&t&-c^2+bc\\ a&b&c\end{vmatrix}=\iff t=b(b-c)$$so $$Q=(a^2:b(b-c):-c^2+bc).$$Note that since $R$ is on cevian $AC$, it can be parameterized by $(1:0:t)$. However, $R,P_\infty,Q$ are colinear, where $P_\infty=(0:-1:1)$ is the point at infinity along $BC$. So we have $$\begin{vmatrix}t&0&1\\ 0&-1&1\\ a^2&b(b-c)&c(b-c)\end{vmatrix}=\iff t=\frac{a^2}{(b+c)(b-c)}$$so $$R=\left(-\frac{a^2}{c^2-b^2}:0:1\right).$$Finally, we just need to check that $R$ lies on the perpendicular bisector of $BC$ which has equation $$0=a^2(z-y)+x(c^2-b^2)$$but this is clearly the case. Done.

This one is really easy... Just use the reim lemma to prove A,P,R,Q are cyclic, and then PR ⊥ BC.