[ [Sum of angles in a triangle. Theorem about the exterior angle.]  [Criteria and properties of an isosceles triangle.]

The lateral sides of the trapezoid are equal to the smaller base, and the diagonals are equal to the larger base. Find the angles of the trapezoid.

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Let $AD > BC$ be the bases of trapezoid $ABCD$, $AC$ and $BD$ be the diagonals, and $AB = CD = BC$ and $AC = BD = AD$.

Let $\angle BAC = \angle ACB = \alpha$. Then

$$
\angle CAD = \angle ACB = \alpha, \angle ACD = \angle ADC = \angle BAD = \alpha + \alpha = 2\alpha
$$

From triangle $CAD$, $2\alpha + 2\alpha + \alpha = 180^\circ$, hence $\alpha = 36^\circ$. Therefore, $\angle BAC = \angle ADC = 2\alpha = 72^\circ$, $\angle ABC = \angle BCD = 108^\circ$.

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-31.jpg?height=179&width=2030&top_left_y=1228&top_left_x=11)

On the sides $AB$ and $BC$ of parallelogram $ABCD$, points $N$ and $M$ are located, respectively, such that $AN:NB = 3:2$, $BM:MC = 2:5$. Lines $AM$ and $DN$ intersect at point $O$. Find the ratios $OM:OA$ and $ON:OD$.

## Hint

Extend $DN$ to intersect line $BC$ and consider two pairs of similar triangles.

## Solution

Extend $DN$ to intersect line $BC$ at point $T$. Let $BM = 6a$, $CM = 15a$. From the similarity of triangles $TNB$ and $DNA$ (coefficient $2/3$), we find that $TB = \frac{2}{3}AD = \frac{2}{3}BC = 14a$, and from the similarity of triangles $TOM$ and $DOA$, $OM:OA = TM:AD = 20a:21a$.

Similarly, we find that $ON:OD = 6:35$.

## Answer

$20:21; 6:35$.

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