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

Labels:
center,
concyclic,
intersecting circles,
radius

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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