The first three conditions are much better expressed like this: two opposite sides of the tetrahedron are equal. From here it follows quickly that $O$ lies on $MM',NN',PP'$, where $M',N',P'$ are the midpoints of $DA,DB,DC$ respectively. Moreover, $MNM'N',NPN'P',PMP'M'$ are rhombi (is this the correct plural of rhombus? ), so the angles we have to prove equal are the angles between the diagonals of three rhombi, which are $\frac \pi 2$.

Call D,E,F,M,N,P the mid point of each side of tetrahedron. So DE, FM, NP have the common midpoint, which is exactly O. Besides, we have DFEN, DNEP, FNMP are lozenges. Thus, <MOP=<MON=<NOP=90