Tricky Triangle

Construction:

Extend AM to point D such that $\overline{AM} = \overline{MD}$ . Thus forming a parallelogram ABDC (This is easy to show using congruent triangles).

Now, in a parallelogram, it is a well-known fact (which is also very easy to prove) that the sum of the squares of the two diagonals is equal to twice the sum of the squares of the two adjacent sides of the parallelogram.

Hence, if $\overline{BC} = x$ , then

$x^2+ (2x)^2 = 2 (3^2 + 4^2)$

therefore,

$5 x^2 = 50$

And finally,

$x = \sqrt{10} = 3.162$

It is a direct application of Stewart's theorem , the proof of which is simple using Law of Cosines (see the proof in Wikipedia).

I used it here, but here's another way I thought of after solving: you could create a parallelogram by adding a triangle (let it be $\triangle BCD$ ) congruent to $\triangle ABC$ with common side $BC$ . Let $\alpha := \angle A$ .

Using law of cosines on $\triangle ABC,\,\triangle ACD$ , knowing $\cos (180-\alpha)=-\cos \alpha$ and $4BC^2=AD^2$ , we get

$4(3^2+4^2-2\cdot 3\cdot 4\cdot \cos \alpha)=4^2+3^2+2\cdot 4\cdot 3\cdot \cos \alpha$

Solve for $\cos \alpha$ , then apply law of cosines on $\triangle ABC$ to find $BC$ .

By Apollonius' theorem

$3^2+4^2=\frac{1}{2} BC^2+ 2AM^2$

As $BC=AM$

$3^2+4^2=\frac{1}{2} BC^2+ 2BC^2$

$25=\frac{5}{2} BC^2$

$\boxed{BC=\sqrt{10}}$