In the quadrilateral $A B C D$, $A B=a, C D=b, A B C \varangle=A D C \varangle=90^{\circ}$ are given. Calculate $A D$, $B C$, and the area of the quadrilateral, assuming that $B A D \varangle$ is an acute angle. Under which conditions is the problem solvable?
Extending the sides $AD$ and $BC$ of the quadrilateral, we get point $E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_59f51478d1e38dc27a39g-1.jpg?height=309&width=333&top_left_y=192&top_left_x=885)

From triangles $ABE$ and $CDE$:

$$
AE = \frac{a}{\cos \alpha}, \quad BE = a \tan \alpha, \quad DE = b \tan \alpha, \quad CE = \frac{b}{\cos \alpha}
$$

Thus,

$$
\begin{aligned}
AD & = AE - DE = \frac{a}{\cos \alpha} - b \tan \alpha = \frac{a - b \sin \alpha}{\cos \alpha} \\
BC & = BE - CE = a \tan \alpha - \frac{b}{\cos \alpha} = \frac{a \sin \alpha - b}{\cos \alpha}
\end{aligned}
$$

The problem is feasible if both $AD$ and $BC$ are positive, that is, if

$$
a > b \sin \alpha \quad \text{and} \quad a \sin \alpha > b
$$

or, equivalently, if

$$
\frac{a}{b} > \sin \alpha > \frac{b}{a}
$$

But $\sin \alpha < \frac{a}{b}$, $\sin \alpha > \frac{b}{a}$.

The area of the quadrilateral is equal to the difference in the areas of the two right triangles. Since the area of triangle $ABE$ is $\frac{a^2 \tan \alpha}{2}$, and the area of triangle $CDE$ is $\frac{b^2 \tan \alpha}{2}$, the area of the quadrilateral is

$$
T = \frac{a^2 - b^2}{2} \tan \alpha
$$

(Bernát Friedmann.)