Proving ABC is Isosceles: Triangle ABC and Bisectors

[jdavel]

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

 

[dodo]

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.

 

[Ben Niehoff]

Have the moderators given up on maintaining the distinction between these forums and the homework forums?

 

[ObsessiveMathsFreak]

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.

 

[jdavel]

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.

 

[Ben Niehoff]

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.

Sorry, my mistake. But there have certainly been a lot of them in this forum lately. It gets annoying after a while.

 

[motx]

If we let \alpha = ABE and \beta = DAB, does then sin\alpha=AE/AB and sin\beta=BD/AB?
And since AE = BD, then \alpha and \beta are equal?

Am I making assumptions from my sketch here?

 

[dodo]

Why sin alfa = AE/AB ? What the Sine Rule says is that (sin alfa) / AE = (sin angle AEB) / AB.

 

[motx]

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.

 

[dodo]

In order to have an hypotenuse, you need a right angle somewhere.

 

[motx]

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.