Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 637.
AG^2=AD^2 - DG^2=(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)so AE^2= AC.AB (1) ( Relation in a right triangle)In circle center D we also haveAG^2= AC.AB (2) ( power of point A to circle D)Compare (1) and (2) we get AE=AGPeter Tran
Using my proof and the result of Problem 636 AG is a tangent to circle CGB, hence AG^2 = AC.ABBut AE^2 also = AC.AB since AE is a tangent to circle BCESo AE = AGSumith PeirisMoratuwaSri Lanka