# 8. Variant 1

In trapezoid $A B C D(A D \| B C)$, the bisectors of angles $D A B$ and $A B C$ intersect on side $C D$. Find $A B$, if $A D=5, B C=2$.
Answer: 7.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_36780c653ec1cfb3ccdag-6.jpg?height=737&width=569&top_left_y=1542&top_left_x=775)

Mark point $L$ on side $AB$ such that $LB = BC$. Let $K$ be the intersection point of the angle bisectors of $\angle DAB$ and $\angle ABC$. Then triangles $LBK$ and $BCK$ are congruent by angle and two adjacent sides $\Rightarrow \angle BLK = \angle BCK$. Therefore, $\angle ALK = 180^\circ - \angle BLK = 180^\circ - \angle BCK = \angle ADC$ (the last equality holds because $AD \parallel BC$). Finally, $\triangle ALK = \triangle ADC$ by angles and a common side. Thus, $AB = AD + BC$.

## Variant 2

In trapezoid $ABCD (AD \parallel BC)$, the angle bisectors of $\angle DAB$ and $\angle ABC$ intersect on side $CD$. Find $AB$ if $AD = 5$, $BC = 3$.

Answer: 8.

## Variant 3

In trapezoid $ABCD (AD \parallel BC)$, the angle bisectors of $\angle DAB$ and $\angle ABC$ intersect on side $CD$. Find $AB$ if $AD = 7$, $BC = 2$.

Answer: 9.

## Variant 4

In trapezoid $ABCD (AD \parallel BC)$, the angle bisectors of $\angle DAB$ and $\angle ABC$ intersect on side $CD$. Find $AB$ if $AD = 11$, $BC = 5$.

Answer: 16.

## Variant 5

In trapezoid $ABCD (AD \parallel BC)$, the angle bisectors of $\angle DAB$ and $\angle ABC$ intersect on side $CD$. Find $AB$ if $AD = 10$, $BC = 3$.

Answer: 13.