SOLUTION: In triangle ABC,m<B = 60 degree. Prove that m( AC)^2=(mAB)^2+(mBC)^2-(mAB).(mCB) ABC is an isosceles triangle in which mAB =mAC.CD is the perpendicular drawn from C to the opp

Algebra ->  Triangles -> SOLUTION: In triangle ABC,m<B = 60 degree. Prove that m( AC)^2=(mAB)^2+(mBC)^2-(mAB).(mCB) ABC is an isosceles triangle in which mAB =mAC.CD is the perpendicular drawn from C to the opp      Log On


   

Question 1067185: In triangle ABC,m m( AC)^2=(mAB)^2+(mBC)^2-(mAB).(mCB)
ABC is an isosceles triangle in which mAB =mAC.CD is the perpendicular drawn from C to the opposite side .Prove that (mBC)^2=2 (mAB).(mBD).
Answer by ikleyn(52803) About Me  (Show Source):
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From the first phrase, cosine of the angle between AB and BC is 1%2F2, hence, the angle ABC is 60 degrees.