Let $ABCD$ be the tetrahedron, $a=[\Delta BCD]$, $b=[\Delta CDA]$, $c=[\Delta DAB]$, $d=[\Delta ABC]$. $h_a,h_b,h_c,h_d$ are the lenght of altitudes from $A,B,C,D$, respectively, and let $r$ be the inradius and $V$ the volume of $ABCD$. Since the midpoint $M_a$ of the height with lenght $h_a$ is on the insphere $\omega$, we have $Vol(M_aBCD) = \frac{V}{2}$, but for all point $P$ in $\omega$ we have that the distance from $P$ to $\Delta BCD$ is at most $2r$, so $Vol(M_aBCD) \le \frac{2ra}{3}$. Therefore, $ar \ge \frac{3V}{4}$, and similarly $br, cr, dr \ge \frac{3V}{4}$ and summing, we have $3V = (a+b+c+d)r \ge 3V$, which implies equality must occur and we conclude that all altitudes passes through the incenter $I$ of $ABCD$. Also, $I$ divides each height in the ratio $3 : 1$, and so $I$ is also the circuncenter or $ABCD$, and its circunradius is $R=3r$. Now, it's not hard to see that $ABCD$ is regular, with each edge having lenght $2\sqrt{6}r$.