[Ratio of areas of triangles with a common angle] Auxiliary similar triangles $\quad]$

In triangle $A B C$, angle $C$ is $30^{\circ}$, and angle $A$ is acute. A line perpendicular to side $B C$ is drawn, cutting off triangle $C N M$ from triangle $A B C$ (point $N$ lies between vertices $B$ and $C$). The areas of triangles $C N M$ and $A B C$ are in the ratio $3: 16$. Segment $M N$ is half the height $B H$ of triangle $A B C$. Find the ratio $A H: H C$.
## Solution

Notice that ${ }^{C N} / B C \cdot C M / A C=S_{M N C} / S_{A B C}=3 / 16$.

Triangle $M N C$ is similar to triangle $B H C$ with a similarity coefficient of $1 / 2$, so ${ }^{C N} /{ }_{C H}=1 / 2,{ }^{C M} /{ }_{C B}=1 / 2$. Therefore, $C N / A C=3 / 8$, and $\quad C H / A C=3 / 4$. Consequently, $A H / H C=1 / 3$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-52.jpg?height=329&width=780&top_left_y=327&top_left_x=639)

## Answer

## Hint

Express the areas of triangles $K B L$ and $B M L$ in terms of the area of parallelogram $A B C D$.

## Solution

$$
\begin{gathered}
S_{\Delta \mathrm{KBL}}=\frac{1}{3} S_{\triangle \mathrm{ABL}}=\frac{1}{3} \cdot \frac{1}{8} S_{\mathrm{ABCD}}=\frac{1}{24} S_{\mathrm{ABCD}} \\
S_{\Delta \mathrm{BML}}=\frac{1}{4} S_{\mathrm{ABCD}}
\end{gathered}
$$

Therefore,

$$
\frac{S_{\triangle K B L}}{S_{\triangle B M L}}=\frac{\frac{1}{24} S_{A B C D}}{\frac{1}{6} S_{A B C D}}=\frac{1}{6}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-52.jpg?height=343&width=828&top_left_y=1861&top_left_x=615)

## Answer

In triangle $A B C$, a line is drawn from point $E$ on side $B C$, parallel to the altitude $B D$, and intersects side $A C$ at point $F$. Segment $E F$ divides triangle $A B C$ into two equal-area figures. Find $E F$, if $B D=6, A D: D C = 2: 7$.

## Hint

$$
S_{C E F / S_{C B A}}={ }^{C E} /{ }_{C B} \cdot{ }^{C F} /{ }_{A C}
$$

## Solution

By the problem's condition, $\frac{1}{2}=\frac{S_{C E F}}{S_{A B C}}=\frac{C E}{B C} \cdot \frac{C F}{A C}=\frac{C E}{B C} \cdot \frac{C F}{\frac{9}{7} C D} \cdot$ Therefore, ${ }^{C E} / C B \cdot C F / C D=9 / 14 \cdot$

By Thales' theorem, $C E: C B=C F: C D$. Thus, $\left({ }^{C E /}{ }^{C B}\right)^{2}=9 / 14, E F=B D \cdot C E /{ }_{C B}=9 \sqrt{\frac{2}{7}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-53.jpg?height=346&width=666&top_left_y=845&top_left_x=702)

## Answer

$9 \sqrt{\frac{2}{7}}$