# Proofs

• 09-04-2003
XSquared
Proofs
i'm just workin on my algebra and discrete geometry homework, and I'm having trouble with the following question:

Quote:

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.
• 09-04-2003
Magos
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.
• 09-04-2003
Silvercord
that's the cosine law
• 09-04-2003
XSquared
Quote:

The bisectors of the exterior angles at A and B meet BC and AC extended at D and E, respectively.
• 09-04-2003
Silvercord
That's what caught me up and why i didn't even attempt to answer

what is a bisector of any angle?
• 09-04-2003
XSquared
A bisector divides the angle in half.
• 09-04-2003
gcn_zelda
hence the word bisect
bi-sect
To divide into two equal or congruent pieces
• 09-04-2003
Perspective
Quote:

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.
• 09-04-2003
JaWiB
So AB == AD == BE, could that mean angle CAB == ABC? Then ACB + 2CAB = 180 ?

Ok just rambling...
• 09-04-2003
XSquared
This is how I interpreted the question: