two tangent lines of parabola with the focus F, tangent in points A & B and meet s point S, create similar triangles.
How can I prove it?
I found a reference to this problem and more:
https://www.cut-the-knot.org/Curricu...aLambert.shtml
I had assumed that $\ell$ was the directrix. However I do not see why $\overline{SA}$ is the perpendicular bisector of $\overline{FA}$. By the definition of parabola we know that $\overline{FA}$ is congruent to $\overline{HA}$. That means that $\Delta FAH$ is isosceles (as well as $\Delta FBG$)
Now from the cut-the-knot I still do not understand the relation of $\mathcal{O}_1$ to $\mathcal{O}_2$
Here's why. Using the picture at cut-the-knot. Extend A'A upward to a point X, FA to the right to a point W, and SA onward to a point Y. Then by the reflective property of the parabola angle FAS = angle XAY. But angle FAS = angle YAW. Therefore angle XAY = angle YAW. Therefore angle FAS = angle SAA' so SA is an altitude of the isosceles triangle FAA' hence the perpendicular bisector of the base.
It means that if the concave side of a parabola was a reflective surface, a beam of light starting at the focus would reflect off the surface and travel parallel to the axis of the parabola (perpendicular to the directrix). That is why spotlights have the shape of a paraboloid with the bulb at the focus.
The reflective property is easily demonstrated using calculus. I don't know if there is a purely geometric argument for it. I suspect not because reflection involves equal angles to a tangent line and the notion of tangent line to a surface is inherently a calculus idea, except for circles.
Note: In my quote there should have been a comma after the word parabola