8,9
[ Angles subtended by equal arcs and equal chords]

Trapezoid $A E F G(E F \| A G)$ is placed in square $A B C D$ with side length 14 such that points $E, F$, and $G$ lie on sides $A B, B C$, and $C D$ respectively. Diagonals $A F$ and $E G$ are perpendicular, and $E G=10 \sqrt{2}$. Find the perimeter of the trapezoid.
Prove that the given trapezoid is isosceles, and the diagonal of the square passes through the point of intersection of its diagonals.

## Solution

Let \( M \) be the point of intersection of the diagonals of the trapezoid, and \( P \) be the projection of point \( E \) onto side \( CD \) of the square. Angles \( \angle BAF \) and \( \angle PEG \) are equal as angles with mutually perpendicular sides. Therefore, the right triangles \( \triangle ABF \) and \( \triangle EPG \) are equal, which means \( AF = EG \). Hence, trapezoid \( AEGF \) is isosceles.

By the Pythagorean theorem, \( BF^2 = AF^2 - AB^2 = 4 \).

Points \( B \) and \( M \) lie on a circle with diameter \( EF \), so \( \angle MBF = \angle MEF = 45^\circ \). Therefore, the diagonal \( BD \) of the square passes through point \( M \).

From the similarity of triangles \( \triangle BMF \) and \( \triangle DMA \), it follows that \( FM : AM = BF : AD = 1 : 7 \). Therefore, \( EF : AG = 1 : 7 \). Let \( EF = x \) and \( AG = 7x \).

Draw the height of the trapezoid \( EH \). Then \( EH = HG = 4x \), \( EG = 4x \sqrt{2} \). Hence, \( x = \frac{5}{2} \).

In triangle \( \triangle AHE \), \( EH = 4x \), \( AG = 3x \), so \( AE = 5x \), and the perimeter of the trapezoid is \( 7x + x + 2 \cdot 5x = 18x = 45 \).

![](https://cdn.mathpix.com/cropped/2024_05_06_8b1cc902b566a3f27ee1g-32.jpg?height=429&width=554&top_left_y=737&top_left_x=755)

## Answer

45.