In the trapezoid of the given figure, $AB$ is parallel to $DC$, $AD=AB=BC=$ $1 \text{ cm}$ and $DC=2 \text{ cm}$. What is the measure of the angle $D \widehat{A} C$?

(a) $30^{\circ}$

(b) $45^{\circ}$

(c) $60^{\circ}$

(d) $90^{\circ}$

(e) $120^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-028.jpg?height=263&width=442&top_left_y=1976&top_left_x=904)
The correct answer is (d).

Let $P$ be the midpoint of segment $D C$ and draw segments $A P$ and $B P$. The three triangles thus formed, $\triangle A D P, \triangle A P B$, and $\triangle B P C$, are equilateral (why?), so $D \widehat{A} P=60^{\circ}=P \widehat{A} B$. Since segment $A C$ is the bisector of angle $P \widehat{A} B$ (why?), we conclude that $P \widehat{A} C=30^{\circ}$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-093.jpg?height=254&width=405&top_left_y=1118&top_left_x=1385)

$$
D \widehat{A} C=D \widehat{A} P+P \widehat{A} C=60^{\circ}+30^{\circ}=90^{\circ}
$$