Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view the complete problem 993
Friday, March 14, 2014
Geometry Problem 993: Intersecting Circles, Secant, Tangent, Concyclic Points, Cyclic Quadrilateral, Perpendicular
Subscribe to:
Post Comments (Atom)
<CBD=(<CBA=<ECD)+(<DBA=<EDC)=180-<CED, which makes CEDB cyclic and all points CEDBF lie on a circle. Then it is obvious that <FBE=<FDE=90.
ReplyDeleteIt also follows that <QFB=<DCB=<AOB/2=<QOB which makes points OQBF lie on a circle.