# Thread: Triangles Where the Sides are Proportional

1. ## Triangles Where the Sides are Proportional

In the course of working a problem related to Menelaus' Theorem, I developed a conjecture which is a new idea to me (I'm sure many other people have thought of it, but I don't know who). It involves triangles formed from 2 segments that cross. Here's a diagram I made in Geogebra: The link below is to the Geogebra file I made, in case anyone wants to move the points around and see the calculations.

https://www.geogebra.org/classic/bqvk6xuy

There's a pair of vertical angles, so the red and blue triangles each have an angle of the same size. Now if I measure the sides, I get

$\displaystyle \frac{{\overline {CA} }}{{\overline {BC} }} \times \frac{{\overline {CE} }}{{\overline {DC} }} = \frac{{\left[ {ACE} \right]}}{{\left[ {BCD} \right]}}$

To put it in words, it seems that the point C where the two segments intersect divides each of them according to a ratio, and the product of those two ratios is equal to the ratio of the areas of the two triangles. The length of the segments $\displaystyle \overline {DB}$ and $\displaystyle \overline {AE}$ seem to have nothing to do with it. The triangles are not similar, and I can move points A, E, B and D around in such a way that these lengths change, but none of the ratios do.

This is something I have observed. I don't know how to prove it, if indeed it is true. My internet searches have not succeeded. I keep getting results about similar triangles, which are involved here.

I hope somebody can help me prove this (if it's true) or else show me why it's false (if that's the case). Thanks!

2. ## Re: Triangles Where the Sides are Proportional

angle of the same size $\displaystyle =\theta$

area ACE $\displaystyle = \frac{1}{2}\overline{\text{AC}}\cdot \overline{\text{CE}} \sin \theta$