4. Let $ABCD$ be an isosceles tangential trapezoid, such that the inscribed circle touches the sides $BC$ and $AD$ at points $L$ and $K$ respectively. Calculate the area of the trapezoid if $\alpha$ is the angle between the diagonals and $\overline{KL}=m$.
Solution. In an isosceles trapezoid with bases $a$ and $b$ and leg $c$, the equality $a+b=2c$ is satisfied. If $E$ is the intersection of the diagonals of the trapezoid and $h_{1}$ and $h_{2}$ are the heights of triangles $ABE$ and $ECD$ dropped from vertex $E$ respectively, then $h_{1}+h_{2}$ is the height of the trapezoid. Let $E_{1} \in AB$ and $E_{2} \in CD$ be the feet of the heights dropped from point $E$.

![](https://cdn.mathpix.com/cropped/2024_06_05_74ce08a8bb971e811c89g-05.jpg?height=343&width=433&top_left_y=410&top_left_x=1127)

If $\beta$ is the angle at vertex $B$ of the trapezoid, then

$$
\sin \beta=\frac{h}{c}
$$

On the other hand, from the right triangles $AE_{1}E$ and $DE_{2}E$ we have $\operatorname{ctg} \frac{\alpha}{2}=\frac{h_{1}}{\frac{a}{2}}$ and $\operatorname{ctg} \frac{\alpha}{2}=\frac{h_{2}}{\frac{b}{2}}$, i.e., $h_{1}=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}$ and $h_{2}=\frac{b}{2} \operatorname{ctg} \frac{\alpha}{2}$, respectively. Therefore,

$$
h_{1}+h_{2}=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}+\frac{b}{2} \operatorname{ctg} \frac{\alpha}{2}=\frac{a+b}{2} \operatorname{ctg} \frac{\alpha}{2}=c \cdot \operatorname{ctg} \frac{\alpha}{2}
$$

From (1) and (2) we get $\sin \beta=\operatorname{ctg} \frac{\alpha}{2}$. If $r$ is the radius of the inscribed circle with center $O$, then from the right triangle $ONL$ we have

$$
\sin \beta=\frac{\frac{m}{2}}{r}=\frac{m}{2 r}=\frac{m}{h}, \text { i.e. } \quad h=\frac{m}{\sin \beta}
$$

Finally,

$$
P=\frac{a+b}{2} h=c \cdot \frac{m}{\sin \beta}=\frac{m}{\operatorname{ctg} \frac{\alpha}{2} \sin \beta} \cdot \frac{m}{\sin \beta}=m^{2} \operatorname{tg}^{3} \frac{\alpha}{2}
$$

## Third Year