[ Theorem of the length of a tangent and a secant; the product of the entire secant and its external part [ Mean proportionals in a right triangle

In rhombus $A B C D$, angle $B A D$ is acute. A circle inscribed in this rhombus touches sides $A B$ and $C D$ at points $M$ and $N$ respectively and intersects segment $C M$ at point $P$, and segment $B N$ at point $Q$. Find the ratio of $B Q$ to $Q N$, if $C P: P M=9: 16$.

#
Apply the Law of Cosines to triangle AMC and the theorem about the tangent and secant.

## Solution

Let $O$ be the center of the circle, $K$ be the point of tangency with side $BC$. Denote $PC=9x, MP=16x, \angle BAC=\alpha, R$ be the radius of the circle. Then

$$
\begin{gathered}
MC=25x, CK^2=CM \cdot CP=25x \cdot 9x=225x^2, CK=15x \\
AM=CK=15x, AO^2=AM^2+OM^2=225x^2+R^2 \\
\cos \alpha=\frac{AM}{OA}=\frac{15x}{\sqrt{225x^2+R^2}}.
\end{gathered}
$$

By the Law of Cosines

$$
MC^2=AM^2+AC^2-2AM \cdot AC \cos \alpha
$$

Substituting $MC, AM, AC$ and $\cos \alpha$ into this equation, we get an equation from which we find that $R=10x$.

Consider the right triangle $AOB$. In it, $MB \cdot MA=MO^2$. Therefore,

$$
MB=\frac{MO^2}{MA}=\frac{R^2}{15x}=\frac{20x}{3}
$$

From the right triangle $BMN$, we find that

$$
BN^2=BM^2+NM^2=\frac{400x^2}{9}+400x^2=\frac{400 \cdot 10x^2}{9}
$$

or

$$
BN=\frac{20x\sqrt{10}}{3}
$$

Since $BQ \cdot BN=MB^2$, then

$$
BQ=\frac{MB^2}{BN}=\frac{2x\sqrt{10}}{3}, QN=BN-BQ=\frac{10x\sqrt{10}}{3}
$$

Therefore, $BQ: QN=1: 9$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-25.jpg?height=437&width=749&top_left_y=-1&top_left_x=660)

## Answer

$1: 9$.