Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 637.
Tuesday, July 19, 2011
Problem 637: Semicircle, Diameter, Perpendicular, Inscribed Circle, Chord, Tangent, Arbelos
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AG^2=AD^2 - DG^2
ReplyDelete=(2R-r)^2 - r^2
=4R^2 - 4Rr
=4R(R-r)
AE^2=AC.AB (tr AEC and tr ABE are similar)
=(2R-2r).(2R)
=4R(R-r)
AE^2=AG^2,
AE=AG
Observe that AEB is a right triangle ( AB is a diameter)
ReplyDeleteso AE^2= AC.AB (1) ( Relation in a right triangle)
In circle center D we also have
AG^2= AC.AB (2) ( power of point A to circle D)
Compare (1) and (2) we get AE=AG
Peter Tran
Using my proof and the result of Problem 636 AG is a tangent to circle CGB, hence AG^2 = AC.AB
ReplyDeleteBut AE^2 also = AC.AB since AE is a tangent to circle BCE
So AE = AG
Sumith Peiris
Moratuwa
Sri Lanka