[ $\left[\begin{array}{l}\text { Auxiliary circle } \\ {[\text { Law of Sines }}\end{array}\right]$

In quadrilateral $A B C D$, it is known that $\angle A B D=\angle A C D=45^{\circ}, \angle B A C=30^{\circ}, B C=1$. Find $A D$.
Prove that points $A, B, C$ and $D$ lie on the same circle and apply the generalized sine theorem ($a=2 R \sin \alpha$).

## Solution

Since the segment $A D$ is seen from points $B$ and $C$, which are on the same side of the line $A D$, under the same angle, the points $A, B, C$ and $D$ lie on the same circle. Let $R$ be the radius of this circle. Then, from triangle $A B C$, we find that

$$
R=\frac{B C}{2 \sin \angle B A C}=\frac{1}{2 \sin 30^{\circ}}=1
$$

and from triangle $A B D$,

$$
A D=2 R \sin \angle A B D=2 \sin 45^{\circ}=\sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-14.jpg?height=434&width=495&top_left_y=0&top_left_x=787)

## Answer

$\sqrt{2}$.