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

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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! 😀

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Best Answer

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.

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