### Instruction: Triangles $ABC$ and $ADC$ are isosceles with $AB=BC$ and $AD=DC$. Point $D$ is inside $\triangle ABC$, $\angle ABC = 40^\circ$, and $\angle ADC = 140^\circ$.  What is the degree measure of $\angle
BAD$?### Response: I notice that $\triangle ABC$ and $\triangle ADC$ are both isosceles, which means that the base angles are congruent.
So, $\angle BAC = \angle BCA = \frac{1}{2}(180^\circ - \angle ABC) = \frac{1}{2}(180^\circ - 40^\circ) = 70^\circ$.
Similarly, $\angle DAC = \angle DCA = \frac{1}{2}(180^\circ - \angle ADC) = \frac{1}{2}(180^\circ - 140^\circ) = 20^\circ$.
I want to find $\angle BAD$, which is the difference between $\angle BAC$ and $\angle DAC$.
Therefore, $\angle BAD = \angle BAC - \angle DAC = 70^\circ - 20^\circ = 50^\circ$.