Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 622.
Sunday, June 12, 2011
Problem 622: Intersecting Circles, Concyclic Points, Centers, Radii
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tr BEC isosceles => ang CEB= ang ECB=a
ReplyDeletetr CAF isosceles => ang CFA= ang ACF=a
thus AEFB is cyclic.
ACBD is a deltoid so ang D = ang C and therefore ang AEB+ ang ADB= ang AEB+ ang ACB=180 => AEBD is cyclic.
finally AEFBD is cyclic.
Tr.s ACB and ADB are congruent hence < ADB = < ACB = < AFB since AC = AF so ADFB is cyclic. Similarly ADBE is cyclic
ReplyDeleteSo AEFBD is cyclic
Sumith Peiris
Moratuwa
Sri Lanka