jmc

By Pythagoras, $\angle ABC=90^{\circ }$. Let $P$ and $Q$ be the projections of $D$ onto $BC$ and $AB$, respectively.



We have that $\angle ABC=90^{\circ }$ and $\angle PBD=45^{\circ }$, so quadrilateral $BPDQ$ is a square. Let $x$ be the side length of this square.

Then $PC=BC-BP=4-x$, and $AQ=AB-QB=3-x$. Triangles $AQD$ and $DPC$ are similar, so $$\frac{AQ}{QD}=\frac{DP}{PC},$$or $$\frac{3-x}{x}=\frac{x}{4-x}.$$Solving for $x$, we find $x=12/7$. Then $BD=x\sqrt{2}=12/7\cdot \sqrt{2}$, so the answer is $\boxed{\frac{12}{7}}$.