Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Thursday, July 12, 2018
Geometry Problem 1361: Triangle, Two Nine-Point Circles, Feuerbach's Circle, Euler's Circle, Congruent Angles, Sketch, iPad Apps
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Triangle ABC is similar to Triangle EBD, Let the centers of circles C1 and C2 are O1 and O2 respectively. By symmetry we have Angle ABO1 = Angle EBO2. Bisector of Angle ABC is also bisector of Angle O1BO2. Let bisector of Angle ABC meets O1O2 at point P. We have
ReplyDeleteBO1/BO2=O1P/O2P=O1G/O2G=O1F/O2F=AB/BE, hence Points B,G,F and P lie on apollonius circle. Also GP=FP as P lies on perpendicular bisector of GF, hence Angle GBP = Angle FBP. Since BP is bisector of Angle ABC we get Angle ABG = Angle CBF