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Geometry ProblemPost your solution in the comment box belowLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 913.
http://img819.imageshack.us/img819/7592/s939.pngDraw F on CD such that CD=DE=eNote that ED perpendicular to CF and triangle CEF is isoscelesIn cyclic quadrilateral BCDE , angle(CED)= angle(CBD)= alpha= angle(DEF)Triangle EAF similar to triangle CEF ………(case AA)So triangle EAF is isosceles => d= B+2d
Peter, don't you need to prove that ED is perpendicular to CD? It's easy to do so but still it needs to be demonstrated, does it not?
AjitYes, By the result of problem 911, BCDE is a cyclic quadrilateral and hence angle CDE = angle CBE= 90Peter.
My comment was quite unnecessary. I didn't see Problem #911 and went straight to 913. Hence the error.
Extend CD to F to form isoceles Tr. ECF with base < 90 - @ and so since < EAF = 2@ EA = AF and the result followsSumith PeirisMoratuwaSri Lanka