In triangle $\triangle ABC$, angle $\angle B$ is equal to $60^\circ$; bisectors $AD$ and $CE$ intersect at point $O$. Prove that $OD=OE$. In triangle $\triangle ABC$, angle $\angle B$ is equal to $60^{\circ}$; bisectors $AD$ and $CE$ intersect at point $O$. Prove that $OD=OE$.
So I've already made a diagram(it is attached below), but I don't know how to prove it from there. Please help and explain your solution thoroughly because I have a test about this tomorrow and I want to understand this! Thank you! :D


Note that O is the incenter of Triangle ABC.
Thus, perpendiculars from O to sides AB and BC have equal length. Let them be G and H respectively. Can you prove that $\triangle$ OGE and $\triangle$ OHD are congruent?
Hint : Use the fact that $\angle$ AOC is $120^o$ and also the fact that $\angle$ GOH is also $120^o$.
