Let this tetrahedron be $ABCD$, and let $M_1$, $M_2$ be the centroids of $\triangle ABC$, $\triangle ACD$, respectively. In addition let $AM_1 \cap BC = E$ and $AM_2 \cap CD = F$. Because we have equal tangents, we can write $AM_1=AM_2$, and therefore $AE=AF$. This also gives us that $\angle EAC = \angle FAC$, so we can write $\triangle EAC \cong \triangle FAC$. This gives, $EC=FC$, and therefore $BC=CD$. We can apply this process repeatedly for different faces of $ABCD$ and get that $AB=AC=AD=BC=BD=CD$, so the tetrahedron is regular.