In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$? In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?  
Alright, so if d is the midpoint then my two triangles are ADB and ADC.
Using the law of cosines I get:
$$3^2=a^2 + 2a^2 - 2\cdot a\cdot 2a\cdot\cos \angle ADB$$ 
for the first triangle,
$$4^2=a^2 + 2a^2 - 2\cdot a\cdot 2a\cdot\cos\angle ADC$$
for the second one.  Then I'm stuck. Any help?

By Stewart's theorem the squared length of the median from $A$ is given by $\frac{2b^2+2c^2-a^2}{4}$.
In our case we know that $c=3, b=4$ and $2b^2+2c^2-a^2 = 4a^2$, hence
$$ a^2 = \frac{2}{5}(b^2+c^2) = 10 $$
and $a=\color{red}{\sqrt{10}}$.
