Answer
m$\angle$DAC=115$^{\circ}$
Work Step by Step
Step one : since the triangle ABC is an isosceles triangle so that the base angles are congruent which means $ 180=m\angle B + \angle A+m \angle B $
By substitution $ 180=50+2\angle A $
So $ \angle A=\angle B= 65 $.
Step two : $ \angle C= \angle BCD + \angle DCA = 65^{\circ} and \angle A= \angle BAD + \angle DAC = 65^{\circ} $
since angle A equal to Angle C and given Angle BCD equal to angle DAC therefore, we conclude that $ \angle DCA= \angle BAD $ .
Step three : by constructing an exterior angle of a triangle ADC and we know that the exterior angle of a triangle equal the sum measures of the two non adjacent interior angles. The exterior angle of triangle ADC is $ 115 + \angle BCD =\angle ADC + \angle DAC $
Since its given that angle BCD is congruent to angle DAC , so by substitution $\angle ADC = 115^{\circ} $.
