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Is there a way to prove the property of the parabola that a ray parallel to its axis of symmetry if "reflected" by the parabola will intersect the symmetry axis of the parabola, without using analytic geometry?

Assuming that $DA\parallel CB$ and that the line has been reflected,I want to prove that B is in fact the focus of the parabola.

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  • $\begingroup$ Idea/Hint: it suffices to show that, if the intersection of the parallel ray with the directrix is $E$, then $\triangle EAB$ is isoceles, and the tangent to the parabola at $A$ either bisects $BE$ or is perpendicular to it (either is sufficient to give two congruent right-angled triangles). $\endgroup$
    – Chappers
    Commented Aug 12, 2017 at 12:34
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    $\begingroup$ There's one at jwilson.coe.uga.edu/EMAT6680Fa08/Wisdom/EMAT6690/Parabolanjw/… , found at physics.stackexchange.com/questions/351498/… $\endgroup$ Commented Aug 12, 2017 at 12:37
  • $\begingroup$ Would you mind a proof that just uses equation of a line ? $\endgroup$
    – user312097
    Commented Aug 12, 2017 at 13:13
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    $\begingroup$ Undermines the point, man. $\endgroup$ Commented Aug 12, 2017 at 15:05

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Let $F$ be focus of parabola.

It is well known that if we project orthogonal any point $P$ on parabola to it directrix we get such point $P'$ on directrix, that perpendicular bisector $t$ of $FP'$ is tangent of parabola at $P$. Thus if we reflect $PG$ across $t$ we get parallel through $P$ to axis of symmetry of parabola.

So if we reverse this process, we see that if reflect parallel to axis of symmetry at point $P$ on parabola, this reflection must go through $F$.

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