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 623.
∠IBE = B/2 + (180°-B)/2 = 90° Similarly ∠ICE = 90°So ∠IBE + ∠ICE = 180° and I,B,E,C are concyclic
internal and external angle bisector are perpendicular; so the result follows
< ECB = 90-C/2 = A/2 + B/2So < AEC = B/2 = < IBC and the result follows Sumith PeirisMoratuwaSri Lanka