The bisectors of the angles B and C of the trapezoid ABCD intersect at the point O lying on the side AD. Prove that the point O is equidistant from the lines AB, BC and CD.
The distance from point O to lines is the length of the perpendicular drawn from the point to the line. In other words, one must prove that ON = OM = OK.
Consider the triangle NBO.
sin∠NBO = ON / OB (by definition of the sine).
ON = OB * sin∠NBO
Consider the triangle BMO.
sin∠OBM = OM / OB (by definition of the sine).
OM = OB * sin∠OBM
∠NBO = ∠OBM (since OB is a bisector).
Therefore, OM = OB * sin∠OBM = OB * sin∠NBO = ON
It is similarly proved that OK = OM.
Those. ON = OM = OK.
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