5. In triangle $A B C$, angle $A$ is twice as large as angle $B . A B=3$ and $A C=2$. Calculate $B C$.
5. Name the intersection of the bisector from $A$ with $B C D$. Then triangle $A B D$ is isosceles.

So $B D = A D$.

Furthermore, it holds that:

$\angle C D A = \angle D B A + \angle D A B = 2 \angle D B A = \angle C A B$.

Thus, triangle CDA is similar to triangle CAB.

So $A B : B C : C A = D A : A C : C D$ or

$3 : B C : 2 = D B : 2 : C D$. Therefore, $3 C D = 2 B D$.

So $D B = \frac{3}{5} B C$. Since $B C \times D B = 6$ we find $B C^2 = 10$ and thus $B C = \sqrt{10}$.
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