This problem was in topic about vectors so i've firstly tried to solve it via cross product formulas but all i've got were just equations with modules that didn't give me anything. Then I've tried to prove it via Heron formulas, but i came to system of 4-th degree equations which i can't solve.
Some solutions via school geometry and projections were found but I can't understand what they are doing. I'm sure it has a solution using vectors. Can anyone give me at least a hint of how to solve this problem?
P.S. Excuses for my English, I don't know it well...
Scale the tetrahedron so two vertices are $A(1,0,0)$ and $B(-1,0,0)$. Rotate the tetrahedron about the line joining these two vertices so the remaining two points are $C(c,y_1,z)$ and $D(d,y_2,z)$.
For equal areas of $ABC$ and $ABD$ we require $y_1^2+z^2=y_2^2+z^2$ and so $y_1=-y_2=y.$
Then twice the area of $ACD$ is given by the modulus of $$\begin{vmatrix}i&j&k \\c-1&y&z\\d-1&-y&z\\\end{vmatrix}=(2yz,(d-c)z,-(c+d-2)y)$$
For $BCD$ we have the same expression but with $-(c+d+2)y$ for the final component.
Therefore $|c+d-2|=|c+d+2|$. Then $c+d=0$ and so $ABC$ and $ABD$ are congruent. This argument could apply to any pair of faces and so the result is proved.