In the diagram, $A D=B D=C D$ and $\angle B C A=40^{\circ}$. The size of $\angle B A C$ is
(A) $80^{\circ}$
(B) $120^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$
(E) $100^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-068.jpg?height=255&width=393&top_left_y=740&top_left_x=1297)
Solution 1

Since $\angle B C A=40^{\circ}$ and $\triangle A D C$ is isosceles with $A D=D C$, then $\angle D A C=\angle A C D=40^{\circ}$.

Since the sum of the angles in a triangle is $180^{\circ}$, then

$\angle A D C=180^{\circ}-\angle D A C-\angle A C D=180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ}$.

Since $\angle A D B$ and $\angle A D C$ are supplementary, then $\angle A D B=180^{\circ}-\angle A D C=180^{\circ}-100^{\circ}=80^{\circ}$.

Since $\triangle A D B$ is isosceles with $A D=D B$, then $\angle B A D=\angle A B D$.

Thus, $\angle B A D=\frac{1}{2}\left(180^{\circ}-\angle A D B\right)=\frac{1}{2}\left(180^{\circ}-80^{\circ}\right)=\frac{1}{2}\left(100^{\circ}\right)=50^{\circ}$.

Therefore, $\angle B A C=\angle B A D+\angle D A C=50^{\circ}+40^{\circ}=90^{\circ}$.

## Solution 2

Since $\triangle A B D$ and $\triangle A C D$ are isosceles, then $\angle B A D=\angle A B D$ and $\angle D A C=\angle A C D$.

These four angles together make up all of the angles of $\triangle A B C$, so their sum is $180^{\circ}$.

Since $\angle B A C$ is half of the the sum of these angles (as it incorporates one angle of each pair), then $\angle B A C=\frac{1}{2}\left(180^{\circ}\right)=90^{\circ}$.

ANSWER: (D)