[The ratio of the areas of triangles with a common base or common height] Complex

On the extension of side $A B$ beyond point $B$ of triangle $A B C$, segment $A D$ is laid out, such that $A D: A B=\alpha$. On the extension of median $B E$, segment $E F$ is laid out, such that $E F: B E=\beta$. Find the ratio of the areas of triangles $B D F$ and $A B C$.

#
Find the ratio of the height $F P$ of triangle $B D F$ to the height $C M$ of triangle $A B C$.

## Solution

Let $P, K$ and $M$ be the projections of points $F, E$ and $C$ onto line $A B$. Then

$$
\frac{P F}{E K}=\frac{B F}{B E}=\beta+1, \frac{E K}{C M}=\frac{1}{2}
$$

Therefore,

$$
\begin{gathered}
S_{\triangle \mathrm{BDF}}=\frac{1}{2} B D \cdot P F=\frac{1}{2}(\alpha-1) A B \cdot \frac{1}{2}(\beta+1) C M= \\
=\frac{1}{2}(\alpha-1)(\beta+1) \cdot \frac{1}{2} A B \cdot C M=\frac{1}{2}(\alpha-1)(\beta+1) S_{\triangle \mathrm{ABC}}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-43.jpg?height=408&width=834&top_left_y=2340&top_left_x=612)

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-44.jpg?height=112&width=2058&top_left_y=224&top_left_x=0)

On the sides of the convex quadrilateral $A B C D$, whose area is 1, points are taken: $K$ on $A B$, $L$ on

$B C$, $M$ on $C D$, $N$ on $A D$. Given that $\frac{A K}{K B}=2, \frac{B L}{L C}=\frac{1}{3}, \frac{C M}{M D}=1, \frac{D N}{N A}=\frac{1}{5}$. Find the area

of the hexagon $A K L C M N$.

## Hint

$S_{\triangle \mathrm{BKL}}=\frac{B K}{A B} \cdot \frac{B L}{B C} \cdot S_{\triangle \mathrm{ABC}}$.

## Solution

Draw diagonal $A C$. Since

$$
\begin{gathered}
S_{\Delta \mathrm{BKL}}=\frac{B K}{A B} \cdot \frac{B L}{B C} \cdot S_{\Delta \mathrm{ABC}}=\frac{1}{12} \cdot S_{\Delta \mathrm{ABC}} \\
S_{\triangle \mathrm{DMN}}=\frac{D N}{A D} \cdot \frac{D M}{D C} \cdot S_{\triangle \mathrm{ADC}}=\frac{1}{12} \cdot S_{\Delta \mathrm{ADC}} \\
S_{\triangle \mathrm{BKL}}+S_{\triangle \mathrm{DMN}}=\frac{1}{12} \cdot\left(S_{\Delta \mathrm{ABC}}+S_{\Delta \mathrm{ADC}}\right)=\frac{1}{12} \cdot S_{\mathrm{ABCD}}=\frac{1}{12}
\end{gathered}
$$

then

$$
S_{\mathrm{AKLCMN}}=S_{\mathrm{ABCD}}-\left(S_{\Delta \mathrm{BKL}}+S_{\Delta \mathrm{DMN}}\right)=1-\frac{1}{12}=\frac{11}{12}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-44.jpg?height=417&width=646&top_left_y=1558&top_left_x=706)

## Answer

$\frac{11}{12}$.