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?

Using $BD=DC=a $, we can write
$$\begin{align} 3^2 &=a^2+(2a)^2-2 \cdot a \cdot 2a \cos(\alpha) \\
&=5a^2-4a^2\cos (\alpha) \end{align}$$
with $\alpha= \angle ADB$.
Also,
$$\begin{align} 4^2&=a^2+(2a)^2-2 \cdot a \cdot 2a\cos (\pi-\alpha) \\
&=5a^2+4a^2\cos (\alpha). \end{align}$$
By addition, we get
$$9+16=25=10a^2. $$
Hence

$$BC = 2a = \sqrt {10}.$$

