# Proofs

This is a discussion on Proofs within the A Brief History of Cprogramming.com forums, part of the Community Boards category; i'm just workin on my algebra and discrete geometry homework, and I'm having trouble with the following question: Triangle ABC ...

1. ## Proofs

i'm just workin on my algebra and discrete geometry homework, and I'm having trouble with the following question:

Triangle ABC is obtuse-angled at C. The bisectors of the exterior angles at A and B meet BC and AC extended at D and E, respectively. If AB = AD = BE, prove that the angle of ACB is 108 degrees.
I'm totally lost as to where to start. If anyone could post a few pointers to get me started, I'd appreciate it.

2. There is a formula (don't know the proper english name for it) that says:

a^2 = b^2 + c^2 - 2*b*c*cos(A)

where a, b & c are the length's of the trinagle's sides, and A is the angle at the opposite side of side a. Implement this in your figure (see below).
You get:

X^2 = (2Y)^2 + (2Y)^2 - 2*(2Y)*(2Y)*cos(a)

=>

X^2 = 8Y^2 - 8Y^2 * cos(a)

=>

X^2 = 8Y^2 * (1 - cos(a))

=>

(X^2)/(8Y^2) = 1 - cos(a)

=>

cos(a) = 1 - (X^2)/(8Y^2)

=>

a = arccos(1 - (X^2)/(8Y^2))

=>

a = arccos(1 - (1/8)*(X/Y)^2)

Now, the problem is to find some kind of relation between X and Y, but I leave that to you.

3. that's the cosine law

4. The bisectors of the exterior angles at A and B meet BC and AC extended at D and E, respectively.

5. That's what caught me up and why i didn't even attempt to answer

what is a bisector of any angle?

6. A bisector divides the angle in half.

7. hence the word bisect
bi-sect
To divide into two equal or congruent pieces

8. Originally posted by XSquared
The bisectors of the exterior angles at A and B meet BC and AC extended at D and E, respectively.
the line that bisects the exterior angle should be that same one that bisects the interior angle. Either im missing some thing in the question or Magos is correct in his diagram.

9. So AB == AD == BE, could that mean angle CAB == ABC? Then ACB + 2CAB = 180 ?

Ok just rambling...

10. This is how I interpreted the question: