A difficult geometry question. (2 Viewers)

[seanieg89]

Prove that any triangle with two internal angle bisectors equal in length is isosceles.

 

[RealiseNothing]

I think I'm missing something in the question because this seems too simple but here it is anyway:

The internal bisectors form an isosceles triangle where the base angles are equal. But since the base angles are bisected, the base angles of the whole triangle must also be equal, hence it's isosceles.

inb4completelywrong.

 

[barbernator]

I think I'm missing something in the question because this seems too simple but here it is anyway:

The internal bisectors form an isosceles triangle where the base angles are equal. But since the base angles are bisected, the base angles of the whole triangle must also be equal, hence it's isosceles.

inb4completelywrong.

inafter, where did you prove that the constructed triangle was isoceles? the two bisected angles cannot be assumed to be same, because that's what we are trying to prove. I also have no idea how to prove it though haha but still trying

 

[RealiseNothing]

inafter, where did you prove that the constructed triangle was isoceles? the two bisected angles cannot be assumed to be same, because that's what we are trying to prove. I also have no idea how to prove it though haha but still trying

I was going on the assumption that the bisectors were equal in length, so the triangle was isosceles, so the base angles are the same.

But now that I think about it, I think we have to prove that the bisectors bisect eachother lol.

 

[barbernator]

question to OP, is this possibly a proof by contradiction?

 

[RealiseNothing]

question to OP, is this possibly a proof by contradiction?

Yes, it's a proof by contradiction.

 

[seanieg89]

Yep, a clever proof by contradiction would be the nicest way of doing it. Alternatively, an ugly coordinate geometry approach will work.

Hint: Standard techniques involving congruence/similarity are surprisingly useless here...try playing with inequalities instead.

 

[barbernator]

Yep, a clever proof by contradiction would be the nicest way of doing it. Alternatively, an ugly coordinate geometry approach will work.

Hint: Standard techniques involving congruence/similarity are surprisingly useless here...try playing with inequalities instead.

haha i cracked and looked up the answer. It is quite a nifty proof but quite long considering how simple the question seems.