[Theorem of the lengths of the tangent and the secant; the product of the entire secant and its external part] [Sine Theorem

A circle passes through vertex $B$ of triangle $A B C$, touches side $A C$ at its midpoint $D$, and intersects sides $A B$ and $B C$ at points $M$ and $N$ respectively, $A B: B C=3: 2$. Find the ratio of the area of triangle $A M D$ to the area of triangle $D N C$.

#
Apply the tangent-secant theorem and the sine rule.

## Solution

From the tangent-secant theorem, it follows that

$$
A M \cdot A B = A D^{2} = C D^{2} = C N \cdot B C
$$

From this, we find that $\frac{A M}{C N} = \frac{B C}{A B} = \frac{2}{3}$. By the sine rule,

$$
\frac{\sin \angle B A C}{\sin \angle A C B} = \frac{B C}{A B} = \frac{2}{3}
$$

$$
\frac{S_{\triangle A M D}}{S_{\triangle D N C}} = \frac{\frac{1}{2} \cdot A M \cdot A D \cdot \sin \angle B A C}{\frac{1}{2} \cdot C N \cdot C D \cdot \sin \angle A C B} = \frac{A M}{C N} \cdot \frac{A D}{C D} \cdot \frac{\sin \angle B A C}{\sin \angle A C B} = \frac{2}{3} \cdot \frac{1}{1} \cdot \frac{2}{3} = \frac{4}{9}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-29.jpg?height=403&width=783&top_left_y=176&top_left_x=635)

## Answer

$4: 9$.