8,9 trapezoids are constructed with circles on their diameters, intersecting at points $M$ and $N$. Chord $M N$ intersects the base $A D$ at point $E$. Find the ratio $A E: E D$.
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The bases of the heights of the trapezoid, dropped from vertices $B$ and $C$, lie on the indicated circles.

## Solution

Let the circle with diameter $B D$ intersect the base $A D$ of the trapezoid $A B C D$ at point $Q$, and the circle with diameter $A C$ at point $P$. Then $C P$ and $B Q$ are the heights of the trapezoid.

Let $C P = B Q = h$. Then $D P = \frac{h}{\sqrt{3}}, A Q = h$. By the theorem of the products of segments of intersecting chords,

$$
D E \cdot E Q = N E \cdot E M = A E \cdot E P, \text{ or } \left(\frac{h}{\sqrt{3}} + P E\right) E Q = (h + Q E) P E
$$

From this, we find that $P E = \frac{E Q}{\sqrt{3}}$. Therefore,

$$
\frac{D E}{A E} = \frac{D P + P E}{A Q + Q E} = \frac{\frac{h}{\sqrt{3}} + \frac{E Q}{\sqrt{3}}}{h + Q E} = \frac{1}{\sqrt{3}}
$$

The condition $A D = 2 B C$ is redundant.

Let the circle with diameter $B D$ intersect the base $A D$ of the trapezoid $A B C D$ at point $Q$, and the circle with diameter $A C$ at point $P$. Then $C P$ and $B Q$ are the heights of the trapezoid.

Let $C P = B Q = h$. Then $D P = \frac{h}{\sqrt{3}}, A Q = h$. By the theorem of the products of segments of intersecting chords,

$$
D E \cdot E Q = N E \cdot E M = A E \cdot E P, \text{ or } \left(\frac{h}{\sqrt{3}} + P E\right) E Q = (h + Q E) P E
$$

From this, we find that $P E = \frac{E Q}{\sqrt{3}}$. Therefore,

$$
\frac{D E}{A E} = \frac{D P + P E}{A Q + Q E} = \frac{\frac{h}{\sqrt{3}} + \frac{E Q}{\sqrt{3}}}{h + Q E} = \frac{1}{\sqrt{3}}
$$

The condition $A D = 2 B C$ is redundant.

Let the circle with diameter $B D$ intersect the base $A D$ of the trapezoid $A B C D$ at point $Q$, and the circle with diameter $A C$ at point $P$. Then $C P$ and $B Q$ are the heights of the trapezoid.

Let $C P = B Q = h$. Then $D P = \frac{h}{\sqrt{3}}, A Q = h$. By the theorem of the products of segments of intersecting chords,

$$
D E \cdot E Q = N E \cdot E M = A E \cdot E P, \text{ or } \left(\frac{h}{\sqrt{3}} + P E\right) E Q = (h + Q E) P E
$$

From this, we find that $P E = \frac{E Q}{\sqrt{3}}$. Therefore,

$$
\frac{D E}{A E} = \frac{D P + P E}{A Q + Q E} = \frac{\frac{h}{\sqrt{3}} + \frac{E Q}{\sqrt{3}}}{h + Q E} = \frac{1}{\sqrt{3}}
$$

The condition $A D = 2 B C$ is redundant.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-20.jpg?height=389&width=540&top_left_y=0&top_left_x=765)

## Answer

$\sqrt{3}$.

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