In triangle ABC, angle C has a measure of 60 degrees. Point D lies on side AC so that BD bisects angle B and BD = 1. Similarly, point E lies on side BC so that AE bisects angle A and AE = 2.
Find the area of triangle ABC.
OOPS  Should not of put the hint in the title.
The length of the medians are
a
AE^2 = bc[1  ()^2]
b+c
and
b
BD^2 = ac[1  ()^2]
a+c
Therefore, AE = 2BD implies
b(b+ca)(a+c)^2 = 4a(a+cb)(b+c)^2 (1)
From the Law of Cosines,
c^2 = a^2+b^22ab*cos(C)
or
c^2 = a^2+b^2ab (2)
Equations (1) and (2) imply (with a lot of algebra)
(4a^2b^2)[c(a+b)+(a^2+b^2)] = 0
This implies
b = 2a
this with equation (2) gives
c = 3*sqrt(3)
Therefore,
b^2 = (2a)^2 = a^2+(a*sqrt(3))^2
= a^2+c^2
Therefore,
B = 90 degrees
Edited on February 12, 2008, 4:13 pm
Edited on February 12, 2008, 4:15 pm
Edited on February 12, 2008, 11:53 pm

Posted by Bractals
on 20080212 16:11:40 