Let $CH$ be the height of the isosceles triangle $ABC$ dropped to the lateral side $AB$, with $AH=2$, $BH=1$. Then $AC=AB=3$.

By the Pythagorean theorem from the right triangles $ACH$ and $BCH$, we find that

$$
CH^2=AC^2-AH^2=9-4=5, BC^2=BH^2+CH^2=1+5=6 \text{. }
$$

Therefore, $BC=\sqrt{6}$.

## Answer

$\sqrt{6}$.
[Pythagorean Theorem (direct and inverse).]

The height of an isosceles triangle, dropped to the lateral side, divides it into segments of 2 and 1, counting from the vertex of the triangle. Find the base of the triangle.