[ Trigonometric ratios in a right triangle [ Theorem of the length of a tangent and a secant; the product of the entire secant and its external part [ Two tangents drawn from the same point

In parallelogram $K L M N$, side $K L$ is equal to 8. A circle, tangent to sides $N K$ and $N M$, passes through point $L$ and intersects sides $K L$ and $M L$ at points $C$ and $D$ respectively. It is known that $K C: L C=4: 5$ and $L D: M D=8: 1$. Find side $K N$.
Apply the tangent-secant theorem.

## Solution

Let $A B$ be the hypotenuse of the right triangle $A B C$. Then $A B=2 R$. If $\angle C A B=\boldsymbol{\alpha}$, then

$$
B C=A B \sin \alpha=2 R \sin \alpha .
$$

If $C D$ is the height of the triangle $A B C$, then

$$
C D=B C \cos \angle D C B=B C \cos \alpha=2 R \sin \alpha \cos \alpha=R \sin 2 \alpha
$$

Let $P$ and $Q$ be the points of tangency of the given circle with the sides $K N$ and $M N$ of the parallelogram $K L M N$. By the tangent-secant theorem,

$$
K P^{2}=K L \cdot K C=8 \cdot \frac{4}{9} \cdot 8=\frac{256}{9} .
$$

Therefore, $K P=\frac{16}{3}$.

Let $K N=M L=x$. Then

$$
Q N=N P=K N-K P=x-\frac{16}{3}, D M=\frac{1}{9} M L=\frac{x}{9}
$$

By the tangent-secant theorem,

$$
M Q^{2}=M D \cdot M L, \text { or }\left(8-x+\frac{16}{3}\right)^{2}=\frac{x^{2}}{9}, \text { or } \frac{40}{3}-x=\frac{x}{3}
$$

From this, we find that $x=10$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-46.jpg?height=426&width=483&top_left_y=2497&top_left_x=793)
[ GMT - circle or arc of a circle ]
Problem