Complete the rectangle ABDF. Join BF to intersect AD at O. AC = BD = AF CB = 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.
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ReplyDeletex=120 Degrees.
ReplyDeleteComplete the rectangle ABDF.
ReplyDeleteJoin BF to intersect AD at O.
AC = BD = AF
CB = 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.
ReplyDeleteHence x = 120
Sumith Peiris
Moratuwa
Sri Lanka