Let $A B C$ be a triangle and suppose $A B=3, B C=4, C A=5$. What is the distance from $B$ to line $A C$ ?
First note that since $3^{2}+4^{2}=5^{2}$, we have $\angle B=90^{\circ}$ by the (converse to the) Pythagorean theorem. We calculate the area of the triangle in two ways. Viewing $A B$ as the base and $B C$ as the height, we get that the area is $\frac{1}{2} \cdot 3 \cdot 4=6$. But viewing $A C$ as the base and $h$ (the distance to be determined) as the height, we get an equation:

![](https://cdn.mathpix.com/cropped/2024_04_17_bf1b51b9b49ddb65304ag-1.jpg?height=288&width=220&top_left_y=295&top_left_x=1641)

$$
\begin{aligned}
\frac{1}{2} \cdot 5 \cdot h & =6 \\
\frac{5}{2} \cdot h & =6 \\
\frac{2}{5} \cdot \frac{5}{2} \cdot h & =6 \cdot \frac{2}{5} \\
h & =\frac{6 \cdot 2}{5}=\frac{12}{5} .
\end{aligned}
$$