8. Variant 1.

On side $A B$ of parallelogram $A B C D$, a point $F$ is chosen, and on the extension of side $B C$ beyond vertex $B$, a point $H$ is chosen such that $A B / B F = B C / B H = 5$. Point $G$ is chosen so that $B F G H$ is a parallelogram. $G D$ intersects $A C$ at point $X$. Find $A X$, if $A C = 100$.
Answer: 40.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_dc91a791fc316ca400e2g-12.jpg?height=588&width=1011&top_left_y=457&top_left_x=571)

In parallelogram $ABCD$, draw diagonal $BD$, and in parallelogram $BFGH$, draw diagonal $GB$. Let $BD$ intersect $AC$ at point $O$. We will prove that $AC \| GB$. Triangles $ABC$ and $GHB$ are similar by two sides and the angle between them: $AB / BF = BC / BH$ and $\angle GHB = \angle ABC$. Therefore, $\angle GBH = \angle ACB$, so $AC \| GB$. Since $O$ is the intersection point of the diagonals of the parallelogram, $DO = BO$, which means $XO$ is the midline of triangle $BDG$. From the similarity, it also follows that $AC / GB = 5$, $GB = 20$. Then $XO = GB / 2 = 10$, $AX = 100 / 2 - 10 = 40$.