1). Ussing barycentric coordonate, if denote by $S_x$ area of the face opose of vertex $X\in\{A,B,C,D\}$, then have: $A(1;0;0;0); B(0;1;0;0); C(0;0;1;0); D(0;0;0;1); O(S_a;S_b;S_c;S_d)$ and if $M_{xy}$ midpoint of edge $[XY]\in\{[BC]; [AD];[AC];[BD]\}$, then: $M_{bc}(0;1;1;0); M_{ad}(1;0;01);M_{ac}(1;0;1;0); M_{bd}(0;1;0;1)$ 2). Or $M_{bc}M_{ac}M_{ad}M_{bd}\mbox{-parallogram}\Rightarrow M_{bd}\in\mbox{plane }(M_{bc}M_{ac}M_{ad})\Rightarrow$ $O\in\mbox{plane }(M_{bc}M_{ac}M_{ad})\Leftrightarrow 0=\left|\begin{array}{cccc} S_a&S_b&S_c&S_d\\ 0&1&1&0\\ 1&0&1&0\\ 1&0&0&1 \end{array}\right|=\\ =\left|\begin{array}{cccc} S_a-S_d&S_b&S_c-S_b&S_d\\ 0&1&0&0\\ 1&0&1&0\\ 0&0&0&1 \end{array}\right|= \left|\begin{array}{cc} S_a-S_d&S_c-S_b\\ 1&1\ \end{array}\right|\Leftrightarrow\\ \Leftrightarrow S_a-S_d=S_c-S_b\Leftrightarrow\boxed{S_a+S_b=S_c+S_d}.$

Let $M,N,L,K$ be the midpoints of $BC,AD,AC,BD.$ Since $MLNK$ is a parallelogram, then $M,L,N,K$ lie on a same plane $\delta.$ It's known that $\delta$ divides the tetrahedron into two pentahedra with equal volumen, hence: $\frac{_1}{^2}[ABCD]=[ODCMK]+[ODCLN]+[OCML]+[OKDN] \pm [OMLNK]$ But $[ODCMK]=\frac{_3}{^4}[OCDB],$ $ [ODCLN]=\frac{_3}{^4}[OCDA],$ $ [OCML]=\frac{_1}{^4}[OABC]$ and $[OKDN]=\frac{_1}{^4}[OABD].$ Thus, substituting the volumen $[ABCD]$ (in the first equation) as the sum of the volumes $[OCDB],$ $[OCDA],$ $[OABC],$ $[OABD],$ yields $[OABC]+[OABD]-[OCDA]-[OCDB]= \pm \ 4 \cdot [OMLNK] \ (\star)$ Since $OABC,OABD,OCDA,OCDB$ have equal O-altitudes (inradius of ABCD), then the given condition is equivalent to $[OABC]+[OABD]=[OCDA]+[OCDB].$ Together with $(\star),$ we get $[OMLNK]=0$ $\Longrightarrow$ $O \in \delta.$