3. As shown in Figure 1, in $\triangle A B C$, point $D$ is on side $B C$. It is known that $A B=A D=2, A C$ $=4$, and $B D: D C=2: 3$. Then $\triangle A B C$ is (
(A) Acute triangle
(B) Right triangle
(C) Obtuse triangle
(D) Cannot be determined
3. (B).

Let $BD=2x$, then $DC=3x$. Draw $AH \perp BD$ at $H$, then $BH=x, CH=4x$. From $2^{2}-x^{2}$. $=AB^{2}-BH^{2}=AH^{2}=$ $AC^{2}-CH^{2}=4^{2}-(4x)^{2}$, solving gives $x^{2}=\frac{4}{5}$.

At this point, $BC^{2}=(5x)^{2}=25 \times \frac{4}{5}=20=2^{2}+4^{2}=$ $AB^{2}+AC^{2}$.