4. Given a regular tetrahedron $A B C D$. Let $E$ be the intersection of the medians of the base $\triangle A B C$, connect $D E$, and take a point $M$ on the line segment $D E$ such that $\angle A M B=90^{\circ}$. Find $E M: M D$.
Since the tetrahedron $ABCD$ is a regular tetrahedron, $D$ is the apex. Let $F$ be the midpoint of edge $AC$ (Figure 10). Without loss of generality, we can assume that the edge length of the tetrahedron $ABCD$ is 1. Therefore,
$$
BF=\frac{\sqrt{3}}{2}, AE=BE=\frac{2}{3} BF=\frac{1}{\sqrt{3}},
$$
$DE=\sqrt{\frac{2}{3}}$. The right triangles $\triangle AEM$ and $\triangle BEM$ are congruent, and $BM=AM$. Given that $\angle AMB=90^{\circ}$, we know that $\triangle AMB$ is an isosceles right triangle, and $AB=BM \cdot \sqrt{2}$. Also, $BM=\sqrt{BE^2+EM^2}$. Thus,
$$
\begin{array}{l}
1=\sqrt{2} \cdot \sqrt{\frac{1}{3}+EM^2}, \\
EM=\frac{1}{\sqrt{6}}=\frac{1}{2} \sqrt{\frac{2}{3}}=\frac{1}{2} DE .
\end{array}
$$

Therefore, $EM: MD=1: 1$.