Let $ABCD$ be a regular tetrahedron with edge $a$, and $M$ be the center of the face $ABC$. Since $DM$ is the height of the tetrahedron, triangle $AMD$ is a right triangle. By the Pythagorean theorem, we find that

$$DM = \sqrt{AD^2 - AM^2} = \sqrt{a^2 - \left(\frac{a \sqrt{3}}{3}\right)^2} = a \sqrt{1 - \frac{1}{3}} = a \sqrt{\frac{2}{3}}$$

The plane passing through the edge $AD$ and the height $DM$ intersects the edge $BC$ at its midpoint $L$. The desired section is the triangle $ADL$ with height $DM$ and base $AL$. Since $DM = a \sqrt{\frac{2}{3}}$ and $AL = \frac{a \sqrt{3}}{2}$, we have

$$S_{\triangle ADL} = \frac{1}{2} AL \cdot DM = \frac{1}{2} \cdot \frac{a \sqrt{3}}{2} \cdot a \sqrt{\frac{2}{3}} = \frac{a^2 \sqrt{2}}{4}$$

## Answer

$\frac{a^2 \sqrt{2}}{4}$.