8,9  [ Area of a triangle (through two sides and the angle between them). ]

In triangle $ABC$, angle $C$ is a right angle, the ratio of the median $CM$ to the bisector $CN$ is $\sqrt{6}$, and the height $CK=2$. Find the areas of triangles $CNK$ and $ABC$.

#
Notice that $C N$ is the bisector of angle $K C M$.

## Solution

Triangles $B M C$ and $A M C$ are isosceles. Therefore, $\angle B C M = \angle B = \angle A C K, \angle K C N = \angle A C N - \angle A C K = \angle B C N - \angle B C M = \angle M C N$, which means $C N$ is the bisector of angle $K C M$.

Let $\angle K C N = \angle M C N = \alpha$. Then $\cos \alpha / \cos 2 \alpha = \sqrt{6}$, from which $\cos \alpha = \sqrt{\frac{2}{3}}, \sin \alpha = \sqrt{\frac{1}{3}}$. Therefore, $C N$

$= C K / \cos \alpha = 2 \sqrt{\frac{3}{2}} = \sqrt{6}$

$A B = 2 C M = 2 C N \sqrt{6} = 12$

Thus, $S_{A B C} = 1 / 2 A B \cdot C K = 12, S_{C N K} = 1 / 2 C N \cdot C K \sin \alpha = \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-51.jpg?height=718&width=211&top_left_y=2065&top_left_x=915)

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-52.jpg?height=217&width=2024&top_left_y=226&top_left_x=11)

Sides $K N$ and $L M$ of trapezoid $K L M N$ are parallel, with $K N = 3$, and angle $M$ is $120^{\circ}$. Lines $L M$ and $M N$ are tangents to the circle circumscribed around triangle $K L N$. Find the area of triangle $K L N$.

## Hint

Prove that triangle $K L N$ is isosceles.

## Solution

Segments $M L$ and $M N$ are equal as tangents drawn from the same point to a circle. Therefore,

$$
\angle M N L = \angle M L N = 30^{\circ}, \angle L N K = \angle M L N = 30^{\circ}.
$$

Since angle $L K N$ is an inscribed angle and angle $M N L$ is the angle between the tangent and chord $L N$, then $\angle L K N = 30^{\circ}$. Therefore, triangle $K L N$ is isosceles. Its height is

$$
\frac{1}{2} K N \cdot \tan 30^{\circ} = \frac{3}{2} \cdot \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{2}
$$

Thus,

$$
S_{\Delta \mathrm{KLN}} = \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{3}}{2} = \frac{3 \sqrt{3}}{4}
$$

## Answer

$\frac{3 \sqrt{3}}{4}$.