In $\triangle ABC$, we have $AB = 1$ and $AC = 2$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$? $\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad$ $\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad$ $\mathrm{(C)}\ \sqrt{2} \qquad$ $\mathrm{(D)}\ \frac 32 \qquad$ $\mathrm{(E)}\ \sqrt{3}$
Solution
Let $D$ be the midpoint of $BC$ and $BD=CD=x$. Then $BC=AD=2x$.
Extend $AD$ to $A'$ with $DA'=AD$; then $ABA'C$ is a parallelogram with diagonals $AA'=4x$ and $BC=2x$. By the parallelogram theorem 26, $AB^2+AC^2+AB^2+AC^2=(2AD)^2+BC^2$, i.e. $10=16x^2+4x^2=20x^2$.
