8-6. Inside triangle $A B C$, a point $D$ is chosen such that $\angle B A D=60^{\circ}$ and $\angle A B C=\angle B C D=30^{\circ}$. It is known that $A B=15$ and $C D=8$. Find the length of segment $A D$. If necessary, round the answer to 0.01 or write the answer as a common fraction
Answer: 3.5.

Solution I. Let the line $A D$ intersect the segment $B C$ at point $X$. Since in triangle $A B X$ the angles $A$ and $B$ are $30^{\circ}$ and $60^{\circ}$, angle $X$ is a right angle. Therefore, $A X = A B / 2 = 7.5$, since $A X$ is the leg of the right triangle $A B X$ opposite the $30^{\circ}$ angle. Similarly, in the right triangle $C D X$, we have $D X = C D / 2 = 4$. Thus, $A D = A X - D X = 3.5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-18.jpg?height=497&width=1308&top_left_y=1285&top_left_x=380)

Solution II. Let the lines $C D$ and $A B$ intersect at point $Y$. The angle $A Y D$ is an exterior angle for triangle $B Y C$, so $\angle A Y D = \angle C B Y + \angle B C Y = 30^{\circ} + 30^{\circ} = 60^{\circ}$. Thus, in triangle $A Y D$, the angles at vertices $A$ and $Y$ are both $60^{\circ}$, making the triangle equilateral. Let $A D = A Y = D Y = x$. Note that triangle $B C Y$ is isosceles because its angles at vertices $B$ and $C$ are equal. Therefore, $15 - x = B Y = C Y = 8 + x$, from which $2 x = 7$ and $x = 3.5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-18.jpg?height=496&width=1308&top_left_y=2262&top_left_x=380)