Task B-3.4. In trapezoid $A B C D$ with perpendicular diagonals, the lengths of the bases $|A B|=a=4,|D C|=c=3$ are known. If the leg $\overline{B C}$ with base $a$ forms an angle of $60^{\circ}$, what is its length?
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_c2de659189058fc448a1g-14.jpg?height=599&width=1179&top_left_y=519&top_left_x=451)

If we draw a parallel to the diagonal $BD$ through vertex $C$ and mark the intersection of the parallel and the extension of side $AB$ as $E$, we get a right triangle $AEC$ whose hypotenuse is of length $a+c$, and the legs are the diagonals $AC$ and $BD (|EC|=|BD|)$. Then, by the Pythagorean theorem, $|AC|^2 + |BD|^2 = (a+c)^2$ or $e^2 + f^2 = 49$.

We will calculate the diagonals by applying the Law of Cosines to triangles $ABC$ and $BCD$. Then we have

$$
4^2 + b^2 - 2 \cdot 4 \cdot b \cos 60^\circ + 3^2 + b^2 - 2 \cdot 3 \cdot b \cos 120^\circ = 49
$$

which gives the quadratic equation

$$
2b^2 - b - 24 = 0
$$

whose positive solution is $b = \frac{1 + \sqrt{193}}{4}$.