Is there a theorem saying that:
If in a parallelogram $ABCD$ the diagonals intersect at $O$, $O$ bisects every segment with ends, lying on $AB$ and $CD$.
I noticed this and I was wondering if I can use it without showing. How can I prove it?
$\triangle AON \cong MOD$.
On
Consider $\triangle OBF$ and $\triangle ODE$. $$OB=OD\\\angle EOD=\angle FOB\\\angle OBF=\angle ODE$$Thus, $\triangle OBF\cong\triangle ODE$. Congruent parts of congruent triangles are equal, which gives you $OF=OE$.
Alternatively, note that $\text{ar}(\triangle AOB)=\text{ar}(\triangle COD)$, a straight-forward consequence of Heron's formula.
$\text{ar}(\triangle AOB)=\frac12AB\times OF'=\frac12AB\times OF\times\cos\angle FOF'\\\text{ar}(\triangle COD)=\frac12CD\times OE'=\frac12AB\times OE\times\cos\angle FOF'$
You can prove this by rotation. Rotate the drawing by half a revolution about the center $O$. Any rotation preserves angles and lengths. Thus $\angle AON$ gets mapped to $\angle DOM$, and $\angle OAB$ gets mapped to $\angle ODC$. Now $\triangle AON$ gets mapped to $\triangle DOM$ which are congruent using ASA congruence since $\overline {DO} \cong \overline {AO}$ . Now $\overline {MO} \cong \overline {NO}$ are congruent because the triangles are congruent. This implies that $O$ bisects $\overline {MN}$.