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?
 A: 
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.
A: Fact $1$ - If we had $UV \parallel BC$ then the line from $A$ bisecting $UV$ would bisect $BC$. (You can prove this using similarity)
With a $\triangle PQR$, drawing tangents to it's circumcircle at $Q$ and $R$ so that they intersect at $X$, $PX$ is known to be the symmedian from point $P$ of $\triangle PQR$. [This is a standard construction of the symmedian, the proof of which you can find at the link above] 
(Symmedian at point $P$ - Median at point $P$ reflected about the internal angle bisector of $\angle QPR$)

This means that $AX$ is the symmedian at point $A$ of $\triangle APQ$.


Note also that $PQ$ is antiparallel to $BC$,

that is same as saying, $(\angle APQ, \angle AQP)=(\angle ACB, \angle ABC)$ (you can prove this using the following chain of arguments $$\angle APQ = \angle ADQ \ \text{(angles in the same segment)} \\ = 180^\circ - \angle AQD - \angle QAD \\ (\because \text{AD is a diameter, angle in semi-circle } \angle AQD = 90^\circ, \\ \text{ and } \triangle ADC \text{ being right-angled }, \angle QAD = 90^\circ-\angle ACB)\\ = 180^\circ - 90^\circ - (90^\circ - \angle ACB)=\angle ACB $$ and similarly for $\angle AQP = \angle ABC$, thus proving that $\triangle AQP$ and $\triangle ABC$ are similar.)
[The definition of antiparallel is a natural extension of the definition of parallel, because if $PQ$ were parallel to $BC$, then we would have had $(\angle AQP, \angle APQ)=(\angle ACB, \angle ABC)$, i.e. the opposite assignment of angles]
Thus, you can view $\triangle APQ$ (similar to $\triangle ABC$) as a smaller version of $\triangle ABC$ with common $\angle A$ which has been reflected about the internal angle bisector at $A$, so that the corresponding side of $AB$ i.e. $AQ$ lies on $AC$ and the corresponding side of $AC$, i.e. $AP$ lies on $AB$. The symmedian for $\triangle APQ$ must then be the median of $\triangle ABC$ (combining the statements in blocks, this follows from fact $1$ stated above).
In particular, the theorem you are looking for is stated here.
A: The problem could be solved immediately using the the main lemma about symmedian.
Lemma
Let $AS$ be symmedian of triangle $ABC$. Then the line $AS$ passes through the intersection of tangents to circumcircle of $ABC$ at points $B$ and $C$.
Proof could be found, for example, in the Wikipedia article linked above, under "Construction of Symmedian".
Solution
By converse of the lemma, $AX$ is a symmedian of triangle $APQ$.
Since $PQ$ and $AB$ are antiparallel, $AX$ is a median of $ABC$. More precisely: by angle chasing we could see that $BC$ becomes parallel to $PQ$ after the symmetry around the bisector of A, thus this symmetry sends median $AM$ of $APQ$ to the median of $ABC$. But by definition of symmedian this symmetry will send $AM$ to $AX$; thus $AX$ contains median  of $ABC$.
A: Pure geometrical solution from a friend of mine with a large math contest background. Appears rather short, although I believe it's complete and correct, but may have some steps omitted.

$AK=KH$, $M_A$ is the midpoint of $BC$.
$H_B,K,H_C,M_A$ lie on the nine point circle $O_9$.
The center of $O_9$ lies on $KM_A$ and bisects $KM_A$ -- the diameter.
$\angle KH_BM_A=\angle KH_CM_A=90^\circ$ $\Rightarrow$
$H_BM_A$ and $H_CM_A$ are tangents to the circle $AH_BHH_C$.
Homothety with coefficient $\frac{AD}{AH}$ centered at $A$ brings $H_B$ to $P$, $H_C$ to $Q$ and $M_A$ to $X$.
