How to solve this question of parabola by only using euclidean geometry

Question

The original question was to prove that $A_1A_2$ and $B_1B_2$ intersect at directrix given $A_1B_1$ and $A_2B_2$ are focal chords of the parabola.

I solved it using the parametric form of parabola and got the answer. But then when I tried using geometry, I'm stuck, in the figure A1B1 is a focal chord, and A2 is a point on parabola, A2A1 intersects directrix at O. Then B2 is the intersection of A1F and B1O,B2' is foot of perp of B2 on directrix. The main target was to prove B2F=B2B'.how am I supposed to prove that the marked angles $x$ in the figure are equal? I'm stuck here!

Answer 1

This theorem is valid not only for a parabola but also for an ellipse or hyperbola:

If $A_1B_1$ and $A_2B_2$ are focal chords of a conic section, then lines $A_1A_2$ and $B_1B_2$ intersect on the directrix referred to that focus.

Let $F$ be a focus of a generic conic section, $A_1B_1$ and $A_2B_2$ two chords through $F$, $A_1'$ and $A_2'$ the projections of $A_1$ and $A_2$ on the directrix, $O$ the intersection between line $A_1A_2$ and the directrix.

From the similitude of triangles $OA_1A_1'$ and $OA_2A_2'$ one gets: $$ OA_1:OA_2=A_1A_1':A_2A_2'=FA_1:FA_2. $$ By the converse of exterior angle bisector theorem we then obtain that line $FO$ is the bisector of external angle $\angle A_1FB_2$ of triangle $FA_1A_2$.

The same reasoning can be repeated for triangle $FB_1B_2$: the line joining focus $F$ with the intersection of line $B_1B_2$ with the directrix is also the bisector of exterior angle $\angle B_2FA_1$. But those exterior angles coincide and have thus the same bisector: it follows that line $B_1B_2$ also intersects the directrix at point $O$. That completes the proof.

EDIT.

The argument given above can be expanded a bit, to prove some interesting properties of the tangents of a conic section.

If we let $A_2\to A_1$ in figure above, then lines $A_1A_2$ and $B_1B_2$ become the lines tangent to the conic at $A_1$ and $B_1$ respectively, endpoints of a focal chord (see figure below). The results proven above still hold, hence those tangents meet at a point $O$ on the directrix, and line $FO$ is the bisector of $\angle A_1FB_1$, i.e. $FO\perp A_1B_1$:

The line tangent to a conic section at a point $P$ meets the directrix relative to focus $F$ at a point $O$ such that $\angle PFO=90°$.

In the particular case of a parabola, it follows that triangles $A_1FO$ and $A_1A_1'O$ are congruent, implying that the line tangent at $A_1$ is the bisector of $\angle A_1'A_1F$, a first well known result:

The line tangent to a parabola at a point $P$ is the bisector of the angle formed by the line passing through $P$ and the focus, with the line through $P$ parallel to the axis.

Finally, from $\angle A_1'A_1F+\angle B_1'B_1F=180°$ one gets $\angle OA_1F+\angle OB_1F=90°$ and consequently $\angle A_1OB_1=90°$, which is another famous property:

The lines tangent to a parabola at the endpoints of a focal chord are perpendicular and meet on the directrix.