Find isosceles trapezoids that are divided by a diagonal into two isosceles triangles.

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Let trapezoid $ABCD$ be divided by diagonal $AC$ into two isosceles triangles, with $AB = CD$. It is easy to see that the equal sides of the isosceles triangles must be: $AB = BC$ and $AC = AD$ (see the figure). $\angle BAC = \angle ACB$ (as the base angles of an isosceles triangle), and $\angle BCA = \angle CAD$ (as alternate interior angles when parallel lines $BC$ and $AD$ are cut by the transversal $AC$). Therefore, $\angle BAC = \angle CAD$, which means diagonal $AC$ is the angle bisector of the base angle of the trapezoid. Let $\angle ACD = \angle ADC = \alpha$. Then $\angle CAD = \angle BAC = 180^\circ - 2\alpha$. On the other hand, $\angle CAD = \alpha / 2$, so $180^\circ - 2\alpha = \alpha / 2$. From this, $\alpha = 72^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_06_198acf192511df7a04ddg-17.jpg?height=300&width=535&top_left_y=1594&top_left_x=767)

## Answer

The sought isosceles trapezoids have a base angle of $72^\circ$, and their diagonals are the angle bisectors of these base angles.