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The ratio in which the bisector divides the side. [The ratio of the areas of triangles with a common base or common height]

In triangle $ABC$, the median $AD$ and the bisector $BE$ are perpendicular and intersect at point $F$. It is known that $S_{DEF}=5$. Find $S_{ABC}$.

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Triangle $A B D$ is isosceles.

## Solution

Triangle $A B D$ is isosceles because its bisector $B F$ is also an altitude. Therefore, $A F=F D \Rightarrow S_{A F E}$

$=S_{D F E}=5$.

Moreover, $B C=2 B D=2 A B$. By the property of the bisector of a triangle, $E C: A E=2: 1$. Therefore, $S_{A B C}=2 S_{A D C}=6 S_{A D E}=60$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-02.jpg?height=269&width=483&top_left_y=117&top_left_x=793)

Answer

60.