On the bases $A D$ and $B C$ of trapezoid $A B C D$, squares $A D E F$ and $B C G H$ are constructed, located outside the trapezoid. The diagonals of the trapezoid intersect at point $O$. Find the length of segment $A D$, if $B C=2, G O=7$, and $G F=18$.
Using the similarity of triangles, prove that points $G, O$, and $F$ lie on the same line.

## Solution

Lines $C G$ and $A F$ are parallel, so $\angle O C G = \angle O A F$. From the similarity of triangles $OBC$ and $ODA$, it follows that $G C: O C = B C: O C = A D: A O = A F: A O$, therefore triangles $G O C$ and $F O A$ are similar. Thus, $\angle G O C = \angle F O A$. Therefore, points $G, O$, and $F$ lie on the same line. Hence, $O F = G F - G O = 18 - 7 = 11$. Consequently, $A D = A F = G C \cdot O F / O G = 22 / 7$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-29.jpg?height=1312&width=426&top_left_y=1&top_left_x=822)

## Answer

$22 / 7$

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