[ Law of Cosines ]

The median $A D$ of an acute-angled triangle $A B C$ is equal to 5. The orthogonal projections of this median on the sides $A B$ and $A C$ are 4 and $2 \sqrt{5}$, respectively. Find the side $B C$.
Find the sines and cosines of the angles formed by the median with sides $A B$ and $A C$, and use the equality of the distances from vertices $B$ and $C$ to the line $A D$.

## Solution

Let $P$ and $Q$ be the projections of point $D$ onto $A B$ and $A C$, and $M$ and $N$ be the projections of points $B$ and $C$ onto the line $A D$. Denote

$$
\angle B A D=\alpha, \angle C A D=\beta, A B=a, A C=b
$$

Then

$$
\cos \alpha=\frac{A P}{A D}=\frac{4}{5}, \sin \alpha=\frac{3}{5}
$$

$$
\begin{gathered}
\cos \beta=\frac{A Q}{A D}=\frac{2}{\sqrt{5}}, \sin \beta=\frac{1}{\sqrt{5}} \\
B M=A B \sin \alpha=\frac{3 a}{5}, C N=A C \sin \beta=\frac{b}{\sqrt{5}}
\end{gathered}
$$

From the equality of the right triangles $C N D$ and $B M D$, it follows that $B M=C N$, i.e., $\frac{3 a}{5}=\frac{b}{\sqrt{5}}$. From this, we find that $b=\frac{3 a}{\sqrt{5}}$.

Expressing the equal segments $B D$ and $C D$ using the cosine theorem in triangles $B A D$ and $C A D$ respectively, we get the equation

$$
a^{2}+25-2 a \cdot 5 \cdot \frac{4}{5}=b^{2}+25-2 b \cdot 5 \cdot \frac{2}{\sqrt{5}}
$$

Substituting $b$ with $\frac{3 a}{\sqrt{5}}$, we get that $a=5$.

Using the cosine theorem in triangle $B A D$, we find that $B D=\sqrt{10}$. Therefore, $B C=2 \sqrt{10}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-33.jpg?height=486&width=763&top_left_y=1052&top_left_x=653)

Answer

$2 \sqrt{10}$.