In the rhombus ABCD O is the point of intersection of the diagonals, OM, OK, OE are the perpendiculars dropped to the sides AB, BC, CD, respectively. Prove that OM = OK and find the sum of the angles MOB and COE.
Consider right-angled triangles BOK and BOM. In these triangles, the hypotenuse OB is common, and the angles <OBK = <OBM, as the angles by which the diagonal of the rhombus divides its angle. And in equal triangles opposite equal angles equal sides are located. Hence, the legs OM and OK are equal, which was required to prove.
Now consider the angles <MOB and <COE. These two angles together with the right angle <BOC are 180 degrees. So the sum of the angles <MOB + <COE = 180 – 90 = 90 (degrees).
One more point: the perpendiculars OM and OE, lowered on the sides of the rhombus AB and CD, represent a common straight line, as perpendiculars lowered on parallel straight lines AB and CD.
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