Given a regular triangle ABC. A semicircle with diameter BC is drawn (It doesn't intersect AB, AC) with 2 points D, E separating it into 3 equal arcs. AD, AE intersects BC at M, N respectively. Prove BM = MN = NC.
I couldn't come up with anything so far...
HINT: If the center of the semicircle is $O$, $\angle BOD = \angle DOE = \angle COE = 60^0$ (same arc lengths will subtend same angle at the center).
So $\triangle ODE$ is equilateral and so segment $DE = R$.
Find altitude of equilateral triangle $AO$. Also find perp from $O$ to $DE$.
Use similar triangles $ADE$ and $AMN$ to find the ratio of $MN$ to $DE$ (knowing ratio of altitudes).
Also as $OM = ON$, you can finally show $BM = MN = NC = \frac{2R}{3}$.