In triangle $\triangle ABC$, angle $\angle B$ is equal to $60^\circ$; bisectors $AD$ and $CE$ intersect at point $O$. Prove that $OD=OE$.

Question

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

Answer 1

Call the angle in $A$ $2\alpha$, then the angle in $C$ is $120-2\alpha$ and $OCA=60-\alpha$. So $DOE=COA=180-(\alpha +60-\alpha)=120$ so the $ODBE$ is cyclic. Now observe that $BO$ must be the bisector of $B$ and so $OED=OBD=30$ and also $ODE=OBE=30$. Since $OE$ and $OD$ are oblique sides of a isosceles triangle they must be equal.

Answer 2

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 $120^o$.

Answer 3

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 $OD = OE$.