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

In rhombus $A B C D$, a circle of radius $R$ is inscribed, touching side $A D$ at point $M$ and intersecting segment $M C$ at point $N$ such that $M N=2 N C$. Find the angles and the area of the rhombus.

#
Apply the tangent-secant theorem and consider the triangle $F P C$, where $P$ and $Q$ are the points of tangency of the inscribed circle with the sides $C D$ and $A B$ of the rhombus $A B C D$.

## Solution

Let $C N=x$. Let $O$ be the intersection point of the diagonals of the rhombus (the center of the inscribed circle), $P$ be the point of tangency of the circle with side $C D$, and $F$ be the point of tangency with side $A B$. Then

$$
C P^{2}=C N \cdot C M=x \cdot 3 x=3 x^{2}
$$

Therefore, $C P=x \sqrt{3}$.

Consider the triangle $F P C$. Its side $P F$ passes through the point $O$,

$$
P F=2 R, F C=C M=3 x, P C=x \sqrt{3}, \angle F P C=90^{\circ} .
$$

By the Pythagorean theorem,

$$
F C^{2}=F P^{2}+P C^{2} \text {, or } 9 x^{2}=4 R^{2}+3 x^{2} .
$$

From this, we find that $x=\frac{R \sqrt{2}}{\sqrt{3}}$.

If $\angle O C P=\alpha$, then

$$
\operatorname{tg} \alpha=\frac{O P}{P C}=\frac{R}{x \sqrt{3}}=\frac{1}{\sqrt{2}}, O C=\frac{O P}{\sin \alpha}=R \sqrt{3}, O D=O C \operatorname{tg} \alpha=\frac{R \sqrt{3}}{\sqrt{2}}
$$

Therefore,

$$
S_{\mathrm{ABCD}}=2 O C \cdot O D=3 R^{2} \sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-33.jpg?height=403&width=532&top_left_y=1462&top_left_x=769)

Answer

$2 \operatorname{arctg} \frac{1}{\sqrt{2}}=\arccos \frac{1}{3} ; 3 R^{2} \sqrt{2}$.