Take a $\sqrt{bc}$-inversion. The problem becomes the following: Quote: Let $\triangle ABC$ be a triangle, and let $P$ be a point such that the midpoint of $\overline{AP}$ lies on $\overline{BC}$. Pick points $X$ and $Y$ on $\overline{AC}$ and $\overline{AB}$ respectively such that \[\angle AXP = \angle ABP,\]and \[\angle AYP = \angle ACP.\]Let $\overline{XY}$ and $\overline{BC}$ meet at $I$. Prove that $\angle API = \angle CAP - \angle BAP$. Now, notice that the pairs of lines $\{\overline{PB}, \overline{PX}\}$ and $\{\overline{PC}, \overline{PY}\}$ are isogonal with respect to $\angle BPX$. Thus, by the isogonality lemma (alternatively, [D]DIT), $\{\overline{PA}, \overline{PI}\}$ are also isogonal with respect to $\angle BPX$. Now, \[\angle API = \angle BPX - 2 \angle APB = \angle BAX - 2 \angle PAB = \angle CAP - \angle BAP,\]so we are done. $\blacksquare$ Remark. Assuming this solution is correct, it doesn't matter $P$ is the midpoint of $AD$?