VII.4. In trapezoid ABCD, diagonal AC is perpendicular to leg BC and is the angle bisector of the angle at vertex A. Calculate the length of the base AB, if $\triangle \mathrm{ABC}=60^{\circ}$, and the perimeter of the trapezoid is $\mathbf{25 cm}$.
Solution: Since AC is the bisector of angle A, we have: $\triangle \mathrm{DAC}=\triangle \mathrm{CAB}$, and $\triangle \mathrm{DCA}=\triangle \mathrm{CAB}$ - as alternate interior angles. It follows that $\triangle D A C=\triangle D C A$, i.e., $\triangle A D C$ is isosceles, $\overline{\mathrm{AD}}=\overline{\mathrm{DC}}$. Triangle $A B C$ is a right triangle and $\triangle C A B=30^{\circ}$, i.e., $\triangle \mathrm{DAB}=60^{\circ}$, so the trapezoid is isosceles.

![](https://cdn.mathpix.com/cropped/2024_06_05_71b0537f301960b82a41g-2.jpg?height=375&width=546&top_left_y=894&top_left_x=914)
The base AB is the hypotenuse in the right triangle $\mathrm{ABC}$, with an angle of $30^{\circ} ; \overline{\mathrm{AB}}=2 \cdot \overline{\mathrm{BC}}$. According to this, $\overline{\mathrm{AD}}=\overline{\mathrm{DC}}=\overline{\mathrm{BC}}=\mathrm{x}$ and $\overline{\mathrm{AB}}=2 \mathrm{x}$, we have $\mathrm{x}+\mathrm{x}+\mathrm{x}+2 \mathrm{x}=25, \mathrm{x}=5 ; \overline{\mathrm{AB}}=2 \cdot 5=10 \mathrm{~cm}$.