Exercise 149

In $\triangle ABC$, $AB=2$ and $AC=3$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?

Solution

With $BD=CD=x$ and median $AD=2x=BC$: by the parallelogram theorem 26, $2(AB^2+AC^2)=(2AD)^2+BC^2=5BC^2$.
$BC^2=\dfrac{2(4+9)}5=\dfrac{26}5$.
$BC=\boxed{\dfrac{\sqrt{130}}5}$.