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Geometry ProblemClick the figure below to see the complete problem 597.
Complete the rectangle ABDF.Join BF to intersect AD at O.AC = BD = AFCB = CD and OB = OD imply CO is the perpendicular bisector of BD and hence of AF as well.Follows AC = FC and triangle ACF is equilateral.So x = supplement of angle CAF = 120 degrees.
Drop perpendiculars CG and CF to AB and BD. Then BGCF is a rectangle and FC= BG=1/2BD=AC. So Tr. FAC is 30-60-90.Hence x = 120Sumith PeirisMoratuwaSri Lanka