[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part

A circle passing through the vertices $B, C$, and $D$ of parallelogram $A B C D$ is tangent to line $A D$ and intersects line $A B$ at points $B$ and $E$. Find the length of segment $A E$, if $A D=4$ and $C E=5$.

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The diagonals of the isosceles trapezoid $B C D E$ are equal. Triangle $B D C$ is isosceles. Next, apply the tangent-secant theorem.

## Solution

Let point $E$ lie between points $A$ and $B$ (Fig.1). The trapezoid $B C D E$ is inscribed in a circle, so it is isosceles. Therefore, $B D = C E = 5$. The chord $B C$ is parallel to the tangent $A D$, so triangle $B D C$ is isosceles (the line passing through point $D$ perpendicular to the tangent $A D$ passes through the center of the circle, is perpendicular to the chord $B C$, and bisects it). Thus, $A B = C D = C E = 5$, and by the tangent-secant theorem, $A D^{2} = A B \cdot A E$, from which $A E = \frac{A D^{2}}{A B} = \frac{16}{5}$. If the point does not lie between points $A$ and $B$ (Fig.2), the problem has no solutions (in this case, reasoning similarly to the first case, we get that $A E = \frac{16}{5} < 5$, which is impossible).

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-29.jpg?height=455&width=535&top_left_y=1825&top_left_x=767)

## Answer

$\frac{16}{5}$.