The circle constructed on side $A D$ of parallelogram $A B C D$ as its diameter passes through the midpoint of diagonal $A C$ and intersects side $A B$ at point $M$. Find the ratio $A M: A B$, if $A C=3 B D$.

#
Quadrilateral $ABCD$ is a rhombus and $\frac{AM}{AB}=\frac{AM}{AD}=\cos \angle BAD$.

## Solution

Let $Q$ be the point of intersection of the diagonals $AC$ and $BD$ of parallelogram $ABCD$. The inscribed angle $AQD$ subtends the diameter $AD$ of the given circle. Therefore, $\angle AQD=90^{\circ}$. Consequently, $ABCD$ is a rhombus and $AB=AD$.

Let $\angle BAD=2\alpha$. Then

$$
\angle QAD=\alpha, \operatorname{tg} \alpha=\frac{DQ}{AQ}=\frac{BD}{AC}=\frac{1}{3}
$$

$$
\cos \angle BAD=\cos 2\alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{4}{5}
$$

Therefore,

$$
\frac{AM}{AB}=\frac{AM}{AD}=\cos 2\alpha=\frac{4}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-24.jpg?height=426&width=992&top_left_y=399&top_left_x=539)

Answer

$4: 5$.