2. In the diagram, point $D$ is on side $B C$ of $\triangle A B C$. If $B D=C D=A D$ and $\angle A C D=40^{\circ}$, what is the measure of $\angle B A C$ ?
2. Since $\triangle A D C$ is isosceles with $A D=D C$, we have $\angle D A C=\angle D C A=40^{\circ}$. Since the sum of the measures of the angles in a triangle is $180^{\circ}$, we have
$$
\angle A D C=180^{\circ}-\angle D A C-\angle D C A=180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ}
$$

Since a straight angle measures $180^{\circ}$, we have $\angle A D B=180^{\circ}-\angle A D C=180^{\circ}-100^{\circ}=80^{\circ}$.
Since $\triangle A D B$ is isosceles with $A D=D B$, we have $\angle D A B=\angle D B A$ and so
$$
\angle D A B=\frac{1}{2}\left(180^{\circ}-\angle A D B\right)=\frac{1}{2}\left(180^{\circ}-80^{\circ}\right)=50^{\circ}
$$

Finally, $\angle B A C=\angle D A B+\angle D A C=50^{\circ}+40^{\circ}=90^{\circ}$.
We note that students who know about "exterior angles" could shorten this solution by noting that, for example, $\angle A D B=\angle D A C+\angle D C A=40^{\circ}+40^{\circ}=80^{\circ}$.
Additionally, some students may note that since $D A=D B=D C$, then $D$ is the centre of the circle that passes through $A, B$ and $C$ (that is, the circumcircle of $\triangle A B C$ ). Since $B C$ will be a diameter of this circle with $A$ a point on its circumference, then $\angle B A C=90^{\circ}$.
ANSWER: $90^{\circ}$