Fixed point of a line through two points on parabola. If $V$ is a vertex (actually it can be any point on parabola) of parabola $\mathcal{P}$ and $A$ and $B$ are
such variable points on $\mathcal{P}$ that $\angle AVB =
90^{\circ}$. Prove that the line $AB$ goes through a fixed point.

I can prove that analyticaly:
Say $y^2 = 4p^2x$, then $V(0,0)$, $A(a^2,2pa)$ and $B(b^2,2pb)$
for some real $a$ and $b$ with condition $4p^2 = -ab$ (because $\angle AVB = 90^{\circ}$). Now the
line $AB$ is $$y ={2p\over a+b}x +{2pab\over a+b}$$ which
intersect the $x$-axis at $x_0=-ab = 4p^2$, which is clearlyconstant.
But how to prove it syntheticaly?
 A: Let $S$ be the focus, $T$ the intersection between line $AB$ and the axis, $H$ and $K$ the projections of $A$ and $B$ on the axis. We want to prove that $VT$ is of fixed length.
We'll repeatedly use Apollonius' definition of parabola, which entails:
$$
AH^2=4VS\cdot VH,\quad BK^2=4VS\cdot VK.
$$
From the similitude of triangles $AHT$, $BKT$ we have $TK:TH=BK:AH$, and 
$(TK+TH):TH=(BK+AH):AH$, hence (supposing WLOG that $VK>VH$):
$$
\begin{align}
TH 
&={TK+TH\over BK+AH}AH={VK-VH\over BK+AH}AH={1\over4VS}{BK^2-AH^2\over BK+AH}AH\\
&={BK-AH\over4VS}AH={BK\cdot AH\over4VS}-{AH^2\over4VS}={BK\cdot AH\over4VS}-VH.\\
\end{align}
$$
It follows that: $\displaystyle VT=TH+VH={BK\cdot AH\over4VS}$.
But from Pythagoras' theorem we also have:
$$
\begin{align}
AB^2
&=(AH+BK)^2+(VK-VH)^2=AH^2+VH^2+BK^2+VK^2+2AH\cdot BK-2VH\cdot VK\\
&=AV^2+BV^2+2AH\cdot BK-2VH\cdot VK=AB^2+2AH\cdot BK-2VH\cdot VK.
\end{align}
$$
It follows that:
$$
AH\cdot BK=VH\cdot VK={1\over16VS^2}AH^2\cdot BK^2,
\quad\text{whence:}\quad
AH\cdot BK=16VS^2.
$$
Substituting that into our previous result for $VT$ gives then $VT=4VS$, and the proof is complete.

