<s>[INST] 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?
 [/INST] You can also practice another method:
Note the areas of $\Delta ABD$ and $\Delta ACD$ are equal. 
Labeling $BD=CD=x$ and $AD=2x$, we will use the Heron's formula:
$$S_{\Delta ABD}= \sqrt{\frac{3+3x}{2}\cdot \left(\frac{3+3x}{2}-3\right)\cdot \left(\frac{3+3x}{2}-2x\right)\cdot \left(\frac{3+3x}{2}-x\right)}=$$
$$S_{\Delta ABD}= \sqrt{\frac{4+3x}{2}\cdot \left(\frac{4+3x}{2}-4\right)\cdot \left(\frac{4+3x}{2}-2x\right)\cdot \left(\frac{4+3x}{2}-x\right)} \Rightarrow$$
$$(3x+3)(3x-3)(3+x)(3-x)=(3x+4)(3x-4)(4-x)(4+x)\Rightarrow$$
$$(9x^2-9)(9-x^2)=(9x^2-16)(16-x^2) \Rightarrow$$
$$81x^2-9x^4-81+9x^2=144x^2-9x^4-256+16x^2 \Rightarrow$$
$$70x^2=175 \Rightarrow x=\sqrt{\frac{5}{2}} \Rightarrow BC=AD=2x=\sqrt{10}.$$
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