3. Four points $A, B, C, D$ are on a plane. It is known that $A B=1, B C=$ $2, C D=\sqrt{3}, \angle A B C=60^{\circ}, \angle B C D=90^{\circ}$. Find $A D$.
Solution. Let's construct the diagram. Let the line $CD$ intersect the line $AB$ at point $O$

![](https://cdn.mathpix.com/cropped/2024_05_06_eefe83f285a2477dfcd6g-1.jpg?height=389&width=868&top_left_y=2127&top_left_x=594)

(according to the condition, these lines are not parallel). There are two possible positions for point $D$ on the line $CD$ - on the segment $CO$ and outside this segment. Triangle $BOC$ is a right triangle, with the angle at vertex $O$ equal to $30^{\circ}$. We find $AB=4$, $OA=3$, $OC=2\sqrt{3}$. For $OD$, considering $CD=\sqrt{3}$, we get two possible values, $OD=\sqrt{3}$ or $OD=3\sqrt{3}$. In triangle $AOD$, two sides and the angle at vertex $O$ are known. Using the cosine rule, we find $AD$. We get two values: $AD=3$ or $AD=\sqrt{3}$. Answer. 3 and $\sqrt{3}$.