Geometry Problem

Click the figure below to see the complete problem 597.

## Friday, April 29, 2011

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## Friday, April 29, 2011

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Geometry Problem 597: Quadrilateral, Right Triangle, Isosceles, Congruence, Angle

## Link List

Online Geometry theorems, problems, solutions, and related topics.

Labels:
congruence,
isosceles,
quadrilateral,
right triangle

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http://www.mathmontada.net/vb/showpost.php?p=20572&postcount=137

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