Apply the Tangent-Secant Theorem.

## Solution

Let $M$ be the midpoint of $CD$. Then $DE = DM = 5$. Let $AK = 2x$, then $KE = 3x$. By the Tangent-Secant Theorem, $KE^2 = BK \cdot AK$, or $9x^2 = (10 + 2x) \cdot 2x$, from which $x = 4$. Therefore, $AK = 8$, $BK = 18$, $KE = 12$, $KD = KE + ED = 12 + 5 = 17$.

By the Pythagorean Theorem, $AD^2 = KD^2 - AK^2 = 17^2 - 8^2 = 15^2$.

Thus, $S_{BCDK} = \frac{1}{2}(18 + 10) \cdot 15 = 210$.

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## Answer

The project is supported by the Department of Education of Moscow and the Federal Targeted Program "Personnel".

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[  [Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part. Pythagorean Theorem (direct and inverse). Area of a trapezoid

A circle passes through vertices $A$ and $B$ of rectangle $A B C D$ and is tangent to side $C D$ at its midpoint. A line is drawn through vertex $D$ that is tangent to the same circle at point $E$, and then intersects the extension of side $A B$ at point $K$.

Find the area of trapezoid $B C D K$, if $A B=10$ and $K E: K A=3: 2$.

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