The title should be "The 3D algebra problem of Shary this year"... The answer is yes. Just write all Pythagorean theorems and then use some times the fact that if $\frac{a}{b}=\frac{c}{d}$, then $\frac{a}{b}=\frac{a \pm c}{b \pm d}$.

Let $AD$ be the altitude of $ABC$. Define $D'$ similarly. Note that $SD\perp BC$ and so $\frac{SB}{SC} = \sqrt{\frac{BD}{DC}} = \sqrt{\frac{B'D'}{D'C'}} = \frac{S'B'}{S'C'}$. Hence, $BC\parallel B'C'$ and hence the other sides are parallel and we are done!