[ Median to the hypotenuse ] [ Area of a triangle (using height and base).]

Outside the right triangle $A B C$ on its legs $A C$ and $B C$, squares $A C D E$ and $B C F G$ are constructed.

The extension of the median $C M$ of triangle $A B C$ intersects the line $D F$ at point $N$. Find the segment $C N$, if $A C=$ $4, B C=1$.
Prove that CN is the height of the right triangle CDF.

## Solution

Right triangles $ABC$ and $DFC$ are equal by two legs, so $DF^{2}=AB^{2}=BC^{2}+AC^{2}=17$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-14.jpg?height=643&width=635&top_left_y=202&top_left_x=723)

The median $CM$ is half the hypotenuse, so $\angle NCF = \angle ACM = \angle A, \angle CNF = 180^{\circ} - \angle NCF - \angle CDF = 180^{\circ} - \angle A - \angle B = 90^{\circ}$, which means $CN$ is the height of the right triangle $CDF$ drawn from the vertex of the right angle $C$. Since $CD \cdot CF = 2S_{CDF} = DF \cdot CN$, then $CN = CD \cdot CF / DF = \frac{4}{\sqrt{17}}$.

## Answer

$\frac{4}{\sqrt{17}}$.