[ Product of lengths of chord segments and lengths of secant segments ] [ Auxiliary similar triangles Area of a triangle (using height and base). ]

Diameter $A B$ and chord $C D$ of a circle intersect at point $E$, and $C E=D E$. Tangents to the circle at points $B$ and $C$ intersect at point $K$. Segments $A K$ and $C E$ intersect at point $M$. Find the area of triangle $C K M$, if $A B=10, A E=1$.
The height of triangle $C K M$ is equal to the segment $B E$. Calculate the tangent of angle $C A E$ and consider similar triangles $А М Е$ and $А К В$.

## Solution

By the theorem of the equality of the products of the segments of intersecting chords, $C E^{2}=9$. Additionally, $C E \perp A B$.

Let $O$ be the center of the circle. Denote $\angle A O C=2 \alpha$. From the isosceles triangle $A O C$, we find that $\angle C A O=90^{\circ}-\alpha$.

Since $\angle B O C=180^{\circ}-2 \alpha$, and $K O$ is the bisector of angle $B O C$, then $\angle B O K=90^{\circ}-\alpha$, and $\angle O K B=\alpha$.

From the right triangle $A E C$, we find that $\operatorname{ctg} \alpha=\operatorname{tg} \angle C A O = \frac{C E}{A E}=3$. Therefore, $K B=O B \operatorname{ctg} \angle O K B = 5 \operatorname{ctg} \alpha = 15$.

![](https://cdn.mathpix.com/cropped/2024_05_06_39480ed8b0d60d88e837g-27.jpg?height=80&width=1630&top_left_y=2770&top_left_x=14)

Thus, $C M = C E - M E = \frac{3}{2}$. Therefore, $S_{C K M} = \frac{1}{2} C M \cdot B E = \frac{27}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_39480ed8b0d60d88e837g-28.jpg?height=751&width=455&top_left_y=-4&top_left_x=810)

## Answer

$\frac{27}{4}$

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