In triangle ABC, angle BAC is 120 degrees. D is on BC so that AD bisects angle BAC. If AB = 2*AC and AD = 100, then find BC.
+128474 Since BAC is a bisected apex angle....then
AB / BD = AC / CD
2AC / BD = AC / CD
2AC /AC = BD/ CD
2 = BD / CD
2CD = BD
And by the Law of Cosines we have that
(BD)^2 = AD^2 + (AB)^2 - 2 (AD*AB)cos (60)
(2CD)^2 = AD^2 + (2AC)^2- 2 (AD * 2AC) (1/2)
4CD^2 = AD^2 + 4AC^2 - 2(AD*AC) (!1)
And
(CD)^2 = AD^2 + AC^2 - 2(AD*AC)cos (60)
CD^2 = AD^2 + AC^2 - (AD*AC) multiply through by 4
4CD^2 = 4AD^2 + 4AC^2 - 4(AD*DC) (2)
Set (1) = (2)
AD^2 + 4AC^2 - 2(AD*DC) = 4AD^2 + 4AC^2 - 4(AD*DC)
3AD^2 - 2(AD*DC) = 0
AD ( 3AD - 2DC) = 0
3AD - 2DC =0
3 (100) = 2DC
DC = 150
And 2CD = BD = 300
So
BC = BD + DC = 300 + 150 = 450
