In an isosceles Delta ABC ,if AB=AC ,BAC=108 and AD divides angle BAC in '2:1' in order then (BD)/(DC) equals to

DeltaABC is an isosceles triangle. AB = AC and BD is perpendicular on AC from vertex B, then BD^(2)-CD^(2) is equal to-

In triangle ABC, if AD perp BC and AD^(2)=BD*DC. Then find the angle BAC

In a Delta ABC it is given that AB=6 cm, AC=8 cm and AD is the bisector of angle A . Then BD:DC=?

In DeltaABC if /_BAC=90^(@) and AB=AC , then /_ABC is

In an acute triangle ABC, /_ABC = 450^@, AB = 3 and AC = sqrt6. The angle /_BAC, is

In Delta ABC,AC=BC and /_ BAC =70^(@) . Find /_BCA

In the given if AD = BD = CD , Find angle BAC