Problem
$O$ is a point in an equilateral triangle $ABC$.
If $∠BAO>∠CAO$, then
$$∠BCO>∠CBO.$$
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Solution
$△ABO$ and $△ACO$ share the side $AO$,
$$AB=AC \qquad and \qquad ∠BAO>∠CAO,$$
$$∴ \ BO>CO,$$
$$∴ \ ∠BCO>∠CBO.$$
$$AB=AC \qquad and \qquad ∠BAO>∠CAO,$$
$$∴ \ BO>CO,$$
$$∴ \ ∠BCO>∠CBO.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.36.