# Thread: similar triangle proof

1. ## similar triangle proof

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?

2. ## Re: similar triangle proof

The triangles are FSB ~ FSA.
(1) The l perpendicular to HA
(2) The l perpendicular to BG
How I can prove that FSB ~ FSA?

3. ## Re: similar triangle proof

What aren't you telling us? What is $I$ and what does it have to do with the problem?

4. ## Re: similar triangle proof

Originally Posted by policer
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?

This one is going to be a bear without any set length scale. I can't think of a way to do this without analytic geometry so we have a whole cluster of variables to deal with. Are you sure there isn't more information?

-Dan

5. ## Re: similar triangle proof

I found a reference to this problem and more:
https://www.cut-the-knot.org/Curricu...aLambert.shtml

6. ## Re: similar triangle proof

Originally Posted by Walagaster
I found a reference to this problem and more:
https://www.cut-the-knot.org/Curricu...aLambert.shtml
I know what to do now. That means we needed to know that I is the directrix of the parabola.

Thanks!

-Dan

7. ## Re: similar triangle proof

Originally Posted by topsquark
I what to do now. That means we needed to know that I is the directrix of the parabola
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 $\mathcal{O}_2$

8. ## Re: similar triangle proof

Originally Posted by Plato
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 $\mathcal{O}_2$
The difference between the proof's diagram and the OPs is rather telling, isn't it? I didn't notice the difference when I said I understood the website's proof.

-Dan

9. ## Re: similar triangle proof

Originally Posted by Plato
I had assumed that $\ell$ was the directrix. However I do not see why $\overline{SA}$ is the perpendicular bisector of $\overline{FA}$
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.

10. ## Re: similar triangle proof

Originally Posted by topsquark
The difference between the proof's diagram and the OPs is rather telling, isn't it? I didn't notice the difference when I said I understood the website's proof.

-Dan
You can move the tangent points and focus around at cut-the-knot so the pictures are quite similar.

11. ## Re: similar triangle proof

Originally Posted by Walagaster
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.
Question: What does the reflective property of the parabola angle mean?

12. ## Re: similar triangle proof

Originally Posted by Plato
Question: What does the reflective property of the parabola, angle mean?
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

13. ## Re: similar triangle proof

Originally Posted by Walagaster
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.
Thank you for that reply. Knowing policer as we do, I assumed that he wanted a synthetic proof. He always wants to know the axioms upon which the properties are based.