Vtyjkyuk

If $V$ is a vertex (actualy it can be any point on parabola) of parabola $mathcalP$ and $A$ and $B$ are
such variable points on $mathcalP$ 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 $$y =2pover a+bx +2pabover a+b$$ which
intersect the $x$-axis at $x_0=-ab = 4p^2$, which is clearly
constant.

But how to prove it syntheticaly?

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:
$$
AH^2=4VScdot VH,quad BK^2=4VScdot VK.
$$
From the similitude of triangles $AHT$, $BKT$ we have $TK:TH=BK:AH$, and
$(TK+TH):TH=(BK+AH):AH$, hence (supposing WLOG that $VK>VH$):
$$
beginalign
TH
&=TK+THover BK+AHAH=VK-VHover BK+AHAH=1over4VSBK^2-AH^2over BK+AHAH\
&=BK-AHover4VSAH=BKcdot AHover4VS-AH^2over4VS=BKcdot AHover4VS-VH.\
endalign
$$
It follows that: $displaystyle VT=TH+VH=BKcdot AHover4VS$.

But from Pythagoras' theorem we also have:
$$
beginalign
AB^2
&=(AH+BK)^2+(VK-VH)^2=AH^2+VH^2+BK^2+VK^2+2AHcdot BK-2VHcdot VK\
&=AV^2+BV^2+2AHcdot BK-2VHcdot VK=AB^2+2AHcdot BK-2VHcdot VK.
endalign
$$
It follows that:
$$
AHcdot BK=VHcdot VK=1over16VS^2AH^2cdot BK^2,
quadtextwhence:quad
AHcdot BK=16VS^2.
$$
Substituting that into our previous result for $VT$ gives then $VT=4VS$, and the proof is complete.