Geometry Problem
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Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 916.
Wednesday, September 4, 2013
Problem 916: Cyclic Quadrilateral, Three Circles, Diameter, Congruence
Labels:
circle,
congruence,
cyclic quadrilateral,
diameter
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draw DF,EA,AC and BD
ReplyDeleteAEB=CFD=ACD=DBA=90
CDA=EBA, BAD=FCD
EBA~CDA =>AD=AB*DC/EB
ABD~CFD =>AD=CD*AB/CF
CD*AB/EB=CD*AB/CF
CF=EB
http://img62.imageshack.us/img62/4753/c84r.png
ReplyDeleteConnect DF and AE and note that DE and AE perpendicular to BC
Let M is the projection of O over BC
We have M is midpoint of BC and EF => BE=CF