In the drawing below, $ABCD$ is a trapezoid and its diagonals $AC$ and $BD$ are perpendicular. Additionally, $BC=10$ and $AD=30$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-12.jpg?height=482&width=762&top_left_y=493&top_left_x=547)

a) Determine the ratio between the segments $BE$ and $ED$.

b) Find the value of the lengths of the segments $EC, AE$, and $ED$ in terms of the length of $BE=x$.

c) If $AE \cdot EC=108$, determine the value of $BE \cdot ED$.
Solution

a) Since $B E$ and $A D$ are parallel, $\angle E B C=\angle E D A$ and $\angle B C E=\angle C A D$. Consequently, triangles $\triangle B E C$ and $\triangle E A D$ are similar, and $B E / E D=B C / A D=10 / 30=1 / 3$. Similarly, we can show that $E C / A E=1 / 3$.

b) From the previous item, we know that $E D=3 B E=3 x$. Using the Pythagorean Theorem in triangle $\triangle B E C$, we have $E C^{2}=B C^{2}-B E^{2}=100-x^{2}$, and consequently $E C=\sqrt{100-x^{2}}$. Since $E C / A E=1 / 3$, it follows that $A E=3 \sqrt{100-x^{2}}$.

c) If $A E \cdot E C=108$, then

$$
108=3 \sqrt{100-x^{2}} \cdot \sqrt{100-x^{2}}=3\left(100-x^{2}\right)
$$

That is, $x^{2}=64$. Therefore,

$$
\begin{aligned}
B E \cdot E D & =x \cdot 3 x \\
& =3 x^{2} \\
& =3 \cdot 64=192
\end{aligned}
$$