They are only the fundamental properties of quasi-tetrahedron, try them in "Problems in solid geometry" of I.F.Sharygin !

gemath wrote: They are only the fundamental properties of quasi-tetrahedron, try them in "Problems in solid geometry" of I.F.Sharygin ! But I haven't got the this book. Can you post your solution?

I change you notation tetrahedron $SABC$ a) With $SA=BC=a,SB=CA=b,SC=AB=c$ in triangle $ABC, bc\cos A=b^{2}+c^{2}-a^{2}$ but easy to see $SA.BC,SB.CA,SC.AB$ is three side of a triangle $\Rightarrow b^{2}+c^{2}>a^{2}\Rightarrow A<\frac{\pi}{2}$ similar to others vetex b)$S_{SAB}=S_{SBC}=S_{SCA}=S_{ABC}$ we will show $SA=BC,SC=AB,SB=AC$ Project $ABCD$ to plane that is parallel with lines $SA,BC$ easily seen the projections of $SBC$ and $ABC$ have the same area similarly projections of $SAB,SAC$ have the same areas Thus the paralelogram with two diagonal projections of $SA,BC$ is projections of $ABCD\Rightarrow SC=AB,SB=AC$ and similarly for other vertex!

gemath wrote: $bc\cos A=b^{2}+c^{2}-a^{2}$ You are wrong! This is solution of a) Let $I$ is midpoint of $BC$, we have $2IA=IA+IS>SA=BC$ therefore $\widehat{BAC}<90^{0}$,...

You are right I forgot number $2$ but in a solution you shouldn't pay some trivial things!