[ Pythagorean Theorem (direct and inverse) ]

In triangle $A B C$, the bisector $B E$ and the median $A D$ are perpendicular and have the same length, equal to 4.

Find the sides of triangle $A B C$.
Triangle $A B D$ is isosceles.

## Solution

Let $P$ be the intersection point of segments $B E$ and $A D$. Triangle $A B D$ is isosceles because its bisector $B P$ is also an altitude. Therefore,

$A P = P D = 2, B C = 2 B D = 2 A B$.

By the property of the bisector of a triangle, $A C = 3 A E$.

Draw a line through vertex $B$ parallel to $A C$. Let $K$ be the intersection point of this line with the extension of median $A D$. Then $B K = A C = 3 A E$.

From the similarity of triangles $A P E$ and $K P B$, it follows that $P E : B P = 1 : 3$.

Therefore, $P E = 1, B P = 3$. Hence, $A B^{2} = A P^{2} + B P^{2} = 13, A E^{2} = A P^{2} + E P^{2} = 5, A C = 3 A E$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3ff818e52eab4ca192bg-06.jpg?height=441&width=692&top_left_y=1686&top_left_x=686)

## Answer

$\sqrt{13}, 2 \sqrt{13}, \sqrt{5}$