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| = 2|BD| implies

   b(b+c-a)(a+c)^2 = 4a(a+c-b)(b+c)^2    (1)

From the Law of Cosines,

   c^2 = a^2+b^2-2ab*cos(C)

       or

   c^2 = a^2+b^2-ab                      (2)

Equations (1) and (2) imply (with a lot of algebra)

   (4a^2-b^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 2008-02-12 16:11:40