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XSquared
09-04-2003, 02:51 PM
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.

Magos
09-04-2003, 03:50 PM
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.

Silvercord
09-04-2003, 04:00 PM
that's the cosine law

XSquared
09-04-2003, 04:02 PM
The bisectors of the exterior angles at A and B meet BC and AC extended at D and E, respectively.

Silvercord
09-04-2003, 04:48 PM
That's what caught me up and why i didn't even attempt to answer

what is a bisector of any angle?

XSquared
09-04-2003, 04:57 PM
A bisector divides the angle in half.

gcn_zelda
09-04-2003, 05:13 PM
hence the word bisect
bi-sect
To divide into two equal or congruent pieces

Perspective
09-04-2003, 05:17 PM
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.

JaWiB
09-04-2003, 05:18 PM
So AB == AD == BE, could that mean angle CAB == ABC? Then ACB + 2CAB = 180 ?

Ok just rambling...

XSquared
09-04-2003, 05:37 PM
This is how I interpreted the question: