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Geometry proof

  1. Jan 15, 2008 #1
    am i missing something, or is this really hard to prove? anyone know a proof? an elegant one?

    Given:
    triangle ABC
    bisector of A intersects BC at D
    bisector of B intersects AC at E
    AE = BD

    Prove: ABC is isosceles
     
  2. jcsd
  3. Jan 15, 2008 #2
    Draw a picture, and apply the Sine Rule like possessed. Notice that the two angles at D have the same sine; this is also true for the two angles at E.
     
  4. Jan 15, 2008 #3

    Ben Niehoff

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    Have the moderators given up on maintaining the distinction between these forums and the homework forums?
     
  5. Jan 15, 2008 #4
    I first encountered this problem almost eight years ago and on occasion still try to prove it. It serves as a kind of meditation, almost Zen in a way.

    I did manage to prove it using coordinate geometry, but of course that's cheating. A paper and compass solution eludes me still. I'm of two minds as to whether I'd like to get a solution from this thread, or still try and solve it myself.
     
  6. Jan 15, 2008 #5
    DoDo,

    That works. Nice!

    P.S. Ben Niehoff, this really wasn't from a homework assignment. I ran across this problem years ago, could never solve it, and just happen to think of it a few days ago.
     
  7. Jan 15, 2008 #6

    Ben Niehoff

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    Sorry, my mistake. But there have certainly been a lot of them in this forum lately. It gets annoying after a while.
     
  8. Jan 17, 2008 #7
    If we let [tex]\alpha[/tex] = ABE and [tex]\beta[/tex] = DAB, does then sin[tex]\alpha[/tex]=AE/AB and sin[tex]\beta[/tex]=BD/AB?
    And since AE = BD, then [tex]\alpha[/tex] and [tex]\beta[/tex] are equal?

    Am I making assumptions from my sketch here?
     
  9. Jan 18, 2008 #8
    Why sin alfa = AE/AB ? What the Sine Rule says is that (sin alfa) / AE = (sin angle AEB) / AB.
     
  10. Jan 18, 2008 #9
    I understand what you're saying, however, all I am using here is sin=opp/hyp. In my construction, sin(beta) = BD/hyp, and sin(alpha)=AE/hyp.

    If they share the same hypotenuse (side AB in my post), then because AE=BE the two angles will be the same.

    Here's my problem: I do believe I assumed AB is the hypotenuse for the two triangles being compared.
     
  11. Jan 18, 2008 #10
    In order to have an hypotenuse, you need a right angle somewhere.
     
  12. Jan 18, 2008 #11
    Right. I realize now the very obvious mistake I made. I chose to draw the triangle equilateral, giving me the right angles I needed. Of course, that breaks the rules from the get-go.
     
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