[ [Ratio of areas of triangles with a common base or common height] Difficult

In a circle, a trapezoid $ABCD$ is inscribed, with its bases $AB=1$ and $DC=2$. Let the point of intersection of the diagonals of this trapezoid be denoted as $F$. Find the ratio of the sum of the areas of triangles $ABF$ and $CDF$ to the sum of the areas of triangles $AFD$ and $BCF$.

#
Let $S$ be the area of triangle $A B F$. Express through $S$ the areas of triangles $C D F, A F D$ and $B F C$.

## Solution

Let $S$ be the area of triangle $A B F$. From the similarity of triangles $A B F$ and $C D F$, it follows that $S_{\triangle \mathrm{CDF}}=4 S$.

Since

$$
\frac{D F}{F B}=\frac{C F}{A F}=\frac{C D}{A B}=2 \text {, }
$$

then

$$
S_{\triangle \mathrm{AFD}}=2 S, S_{\triangle \mathrm{BFC}}=2 S .
$$

$$
\frac{S_{\triangle A B F}+S_{\triangle C D F}}{S_{\triangle A F D}+S_{\triangle B F C}}=\frac{S+4 S}{2 S+2 S}=\frac{5}{4}
$$

This statement is true for any trapezoid where one base is twice the length of the other.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-36.jpg?height=395&width=458&top_left_y=214&top_left_x=800)

## Answer

$\frac{5}{4}$.