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

Each diagonal of the convex pentagon ABCDE cuts off a triangle of unit area from it. Calculate the area of the pentagon ABCDE

#
Prove that the diagonals of the given pentagon are parallel to its sides.

## Solution

Let $K$ be the point of intersection of the diagonals $AC$ and $BE$. Since $S_{\triangle ABE} = S_{\triangle ABC}$, and $S_{\triangle AKE} = S_{\triangle BKC}$. Therefore,

$AK \cdot KE = BK \cdot KC$, or $\frac{AK}{KC} = \frac{BK}{KE}$. Hence, $EC \parallel AB$. Similarly, we can prove that the other diagonals are also parallel to the corresponding sides.

Since $DEKC$ is a parallelogram, then $S_{\triangle EKC} = S_{\triangle EDC} = 1$. Let $S_{\triangle AKE} = x$. Then

$$
\frac{S_{\triangle AKE}}{S_{\triangle AKB}} = \frac{EK}{KB} = \frac{x}{1-x} = \frac{S_{\triangle EKC}}{S_{\triangle CKB}} = \frac{1}{x}
$$

From the equation $\frac{x}{1-x} = \frac{1}{x}$, we find that $x = \frac{\sqrt{5} - 1}{2}$. Therefore,

$$
S_{ABCDE} = S_{\triangle ABC} + S_{\triangle DEKC} + S_{\triangle AKE} = 1 + 2 + x = \frac{5 + \sqrt{5}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f7c9453eb6f3f37ea7g-42.jpg?height=429&width=712&top_left_y=2244&top_left_x=679)

## Answer

## Problem