Let $ABCD$ be the given trapezoid, $AB=a, DC=c, \angle DAB=A, \angle DCB=C, AD=d, CB=b$. From $C$, draw a line parallel to $AD$ intersecting $AB$ at $E$. In $\triangle ECB$, $EB=a-c=m, \angle ECB=\gamma-\alpha=E, \angle CEB=\alpha$ and $\angle EBC=180^{\circ}-\gamma$. Applying the Mollweide's formula to this triangle:

$$
\frac{d-b}{a-c}=\frac{\sin \frac{1}{2}\left(180^{\circ}-\gamma-\alpha\right)}{\cos E}
$$

or

$$
\frac{d-b}{a-c}=\frac{\cos \frac{\gamma+\alpha}{2}}{\cos E}
$$

from this equation we calculate $(\alpha+\gamma)$, and since $(\alpha-\gamma)$ is known, $\alpha$ and $\gamma$ can be determined. -Then, from $\triangle CEB$, we determine $d$; drawing a perpendicular from $C$ to $AB$, we obtain the height $CF$ of the trapezoid, which we calculate from $\triangle CEF$; but the area of the trapezoid is:

$$
t=\frac{a+c}{2} \cdot CF
$$

and thus $a+c$ can be calculated. Substituting the given values, we find that $a+c=34 \mathrm{~s}$, and since $a-c=14$, therefore $a=24$ $\mathrm{cm}$ and $c=10 \mathrm{~cm}$.

The problem was solved by: Freibauer E., Kiss A., Krausz B., Krisztián Gy., Lukhaub Gy., Weisz J.
The area of a trapezoid is $t=204 \mathrm{~m}^{2}$; the difference between the parallel sides is $a-c=m=14 \mathrm{~m}$, the difference between the non-parallel sides is $b-d=n=2 \mathrm{~m}$, and the difference between the opposite angles is $C-A=E=59^{\circ} 29^{\prime} 23^{\prime \prime}$. Calculate the parallel sides of the trapezoid.