Median of triangle and tangents Let $AD$ be an altitude of any triangle $ABC$. Consider a circle with $AD$ as its diameter, cutting $AB$ and $AC$ at $P$ and $Q$ respectively. Let the tangents at $P$ and $Q$ meet at $X$. Prove that $AX$ bisects $BC$.
I have attempted the proof, using angles in the alternate segment, and Apollonius’ Theorem, but to no avail. Are there some other theorems that I should be using?


Pure coordinate proof:
$D:=(0,0),\,A:=(0,a),\,B:=(b,0),\,C:=(c,0),$
$O:=\left(0,\frac{a}{2}\right),\,P:=pB+(1-p)A,\,
Q:=qC+(1-q)A,\,X:=(x,y)$
$$\begin{cases}
OP^2=\frac{a^2}{4}\\
OQ^2=\frac{a^2}{4}\\
\overrightarrow{OP}\cdot\overrightarrow{XP}=0\\
\overrightarrow{OQ}\cdot\overrightarrow{XQ}=0\\
\end{cases}$$
$$\begin{cases}
\left(pb,\left(\frac12-p\right)a\right)^2=\frac{a^2}{4}\\
\left(qc,\left(\frac12-q\right)a\right)^2=\frac{a^2}{4}\\
\left(pb,\left(\frac12-p\right)a\right)\cdot
\left(pb-x,\left(1-p\right)a-y\right)=0\\
\left(qc,\left(\frac12-q\right)a\right)\cdot
\left(qc-x,\left(1-q\right)a-y\right)=0\\
\end{cases}$$
$$\begin{cases}
p^2b^2+\frac{a^2}{4}-pa^2+p^2a^2=\frac{a^2}{4}\\
q^2c^2+\frac{a^2}{4}-qa^2+q^2a^2=\frac{a^2}{4}\\
p^2b^2-pbx+\frac{a^2}{2}-pa^2-\frac{pa^2}{2}+p^2a^2
-\frac{ay}{2}+pay=0\\
q^2c^2-qcx+\frac{a^2}{2}-qa^2-\frac{qa^2}{2}+q^2a^2
-\frac{ay}{2}+qay=0\\
\end{cases}$$
$$\begin{cases}
pb^2-a^2+pa^2=0\hbox{, as }p\ne 0\ (p=0\hbox{ gives }P=A),\\
qc^2-a^2+qa^2=0\hbox{, as }q\ne 0\ (q=0\hbox{ gives }Q=A),\\
-2pbx+(-a+2pa)y=-2p^2b^2-a^2+3pa^2-2p^2a^2\\
-2qcx+(-a+2qa)y=-2q^2c^2-a^2+3qa^2-2q^2a^2\\
\end{cases}$$
$$\begin{cases}
p=\frac{a^2}{a^2+b^2}\\
q=\frac{a^2}{a^2+c^2}\\
-2pbx+(-a+2pa)y=-2p^2b^2-a^2+3pa^2-2p^2a^2\\
-2qcx+(-a+2qa)y=-2q^2c^2-a^2+3qa^2-2q^2a^2\\
\end{cases}$$
From the 3ed and the 4th equations we find $x,y$ by elimination or by Cramer's rule, it appears rather cumbersome, so I'll jump to the result:
$$x = \frac{a^2(b + c)}{2(a^2 + bc)},\, y = \frac{abc}{a^2 + bc},$$
so it's only left to show that $\overrightarrow{AX}=t\left(\frac{B+C}{2}-A\right)$ for some $t$, and indeed after comparing expressions for $\overrightarrow{AX}=X-A$ and $\left(\frac{b+c}{2},-a\right)$ we find $t=\frac{a^2}{a^2 + bc}$, QED.
