In the trapezoid in the figure below, $AB$ is parallel to $CD$, $AD = AB = BC = 1 \, \text{cm}$, and $DC = 2 \, \text{cm}$. What is the measure of the angle $\angle CAD$?

(A) $30^{\circ}$

(B) $45^{\circ}$

(C) $60^{\circ}$

(D) $90^{\circ}$

(E) $120^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_01_eead07372ad548e635bag-02.jpg?height=241&width=398&top_left_y=720&top_left_x=950)
The correct answer is (D).

Let $P$ be the midpoint of segment $C D$ and draw segments $A P$ and $B P$. The three triangles formed $\triangle A D P$, $\triangle A B P$, and $\triangle B C P$ are equilateral (why?). Therefore, the angles $D \widehat{A} P=60^{\circ}=P \widehat{A B}$. 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_eead07372ad548e635bag-13.jpg?height=244&width=401&top_left_y=1165&top_left_x=1339)

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