Results 1 to 2 of 2

Thread: Triangles Where the Sides are Proportional

  1. #1
    Member
    Joined
    Nov 2012
    From
    Normal, IL USA
    Posts
    197
    Thanks
    29

    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:


    Triangles Where the Sides are Proportional-p5-problem-2-4-ratio-areas.png
    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,039
    Thanks
    530

    Re: Triangles Where the Sides are Proportional

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

    area ACE $\displaystyle = \frac{1}{2}\overline{\text{AC}}\cdot \overline{\text{CE}} \sin \theta$
    Last edited by Idea; Aug 26th 2019 at 10:34 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Two triangles connected together ( Find four sides)
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Mar 31st 2016, 03:36 PM
  2. Replies: 2
    Last Post: Dec 18th 2012, 03:25 AM
  3. Ratios of sides without similar triangles
    Posted in the Geometry Forum
    Replies: 0
    Last Post: Mar 29th 2012, 09:19 PM
  4. In the right triangles with sides...
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Apr 21st 2010, 05:43 AM
  5. Sides of Congruent Triangles
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Jun 18th 2008, 06:29 PM

/mathhelpforum @mathhelpforum