jmc

Let us draw a diagram first: As we have isosceles triangles $△ABD$ and $△ACD,$ let us have $\angle ABC=a$ and $\angle ACB=b.$ Then, $\angle BAC=\angle BAD+\angle DAC=a+b.$ Since the three angles must sum to $180^{\circ },$ we have that $2(a+b)=180^{\circ },$ so $a+b=90^{\circ }.$ Since we know that $a=50^{\circ },$ we see that $b=\angle ACB=\boxed{40^{\circ }}.$