jmc

Let $\angle CAD=\angle BAD=x$, and let $\angle ACD=\angle BCD=y$ From $△ABC$, we know that $50+2x+2y=180$, leading to $x+y=65$. From $△ADC$, we know that $x+y+\angle D=180$. Plugging in $x+y=65$, we get $\angle D=180-65=115$, which is answer $\boxed{115^{\circ }}$.