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

In triangle $ABC$, a line is drawn from vertex $A$, intersecting side $BC$ at point $D$, which lies between points $B$ and $C$, and $\frac{CD}{BC}=\alpha\left(\alpha<\frac{1}{2}\right)$. On side $BC$ between points $B$ and $D$, a point $E$ is taken, and a line is drawn through it, parallel to side $AC$, intersecting side $AB$ at point $F$. Find the ratio of the areas of trapezoid $ACEF$ and triangle $ADC$, given that $CD=DE$.

#
Express the areas of trapezoid $A C E F$ and triangle $A D C$ in terms of the area of triangle $A B C$.

## Solution

Let $B C=x$. Then

$$
\begin{gathered}
C D=\alpha x, D E=C D=\alpha x, C E=2 \alpha x, B E=(1-2 \alpha) x \\
S_{\Delta \mathrm{FBE}}=(1-2 \alpha)^{2} S_{\Delta \mathrm{ABC}} \\
S_{\mathrm{ACEF}}=\left(1-(1-2 \alpha)^{2}\right) S_{\Delta \mathrm{ABC}}=4 \alpha(1-\alpha) S_{\Delta \mathrm{ABC}} \\
S_{\Delta \mathrm{ADC}}=\frac{C D}{B C} S_{\Delta \mathrm{ABC}}=\alpha S_{\Delta \mathrm{ABC}}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-58.jpg?height=441&width=778&top_left_y=154&top_left_x=640)

## Answer

$4(1-\alpha)$.

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