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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 874.
Let the mid point of O1O2 be M, M must lies on BD (the radical axis of two circles)Angle(BO1O2)=angle(BO1M)=angle(BAD)=angle(BCD)=angle(BO2M)=angle(BO2O1)So BO1O2 forms an isosceles triangleQ.E.D.
build:O2DO2BO1BO1DA=C=aBO2D=2a=BO1DO2BD=O2DB=O1BD=O1DB=90-aO1BO2=O1DO2=180-2aBO1DO2 parallelogramO2D=O1Bcircumcircle O1 and O2 are congruentQ.E.D.
2×BO1 = BD / sin∠BAC = BD / sin∠BCA = 2×BO2The result follows.
< BO1D = 2A hence < ABO1 = 90-A - < ABD Similarly < CBO2,= 180 - 2A - < ABD - (90-A) = < ABO1So Tr.s ABO1 and BCO2 are congruent ASA and the result followsSumith PeirisMoratuwaSri Lanka
It also follows that BO1DO2 is a rhombus so BD and O1O2 bisect each other perpendicularly