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
$$\begin{aligned}\angle DOE &= \angle AOC\\ &= 180^{\circ} - (\angle OAC + \angle OCA)\\ &= 180^{\circ} - \frac{1}{2} (\angle BAC + \angle BCA)\\ &= 180^{\circ} - \frac{1}{2} (180^{\circ} - \angle ABC)\\ &= 120^{\circ}.\end{aligned}$$
Hence, $\angle DOE + \angle BOE = 180^{\circ}$, so $BDOE$ is cyclic. Therefore, since $\angle ODE = \angle OBE = \angle OBD = \angle OED$, $OED$ is an isosceles triangle, so $OD = OE$.
