Question

ABC IS A TRIANGLE IN WHICH ANGLE B=2 OF ANGLE C.D IS A POINT ON BC SUCH THAT AD BISECTS ANGLE BAC AND AB=CD .PROVE THAT ANGLE BAC =72DEGREE

Answer 1

Constr:draw angle b from angle e solution :let angle a=2y                      angle b=2x      Angle BAD=angle CAD=y     Angle ABE=angle CBE=x  In triangle BCE  Angle EBC=angle ECB=angle x EC=BE (side opp.to equal angle )  In triangle ABE and triangle DCE AB=CD (given) angle ABE=angle DCE (x) BE=CE (above proved ) triangle ABE congruent to triangle DCE (S.A.S) DE=AE (C.P.C.T)  in triangle AED angle EAD=angle EDA=y (angle opp.to equal side) angle CED=angle AEB (C.P.C.T) angle CED=y+y (exterior angle property of triangle AED) angle CED=2y  angle BEA=x+x(exterior angle of triangle BEC) angle BEA=2x  So,2x=2y  In triangle ABC angle A+angle B+angle C=180° 2y+2x+x=180°(2y=2xand x=y) 2y+2y+y=180° 5y=180° y=36°  angle BAC=2y=2×36=72°                                        proved