Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Friday, July 17, 2015
Geometry Problem 1139: Triangle, Circumcircle, Angle Bisector, Parallel Lines, 90 Degrees, Concurrent Lines
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Line FG meets Circle O at point K,
ReplyDeleteAngle BKG = Angle C/2
Also in Triangle GHB Angle BHG = Angle C/2
So Quad BGKH is cyclic
Angle HGB = Angle A/2 = Angle HKB
Since Angle DKB = Angle A/2
HDK must be collinear
Similarly JEK must be collinear
it proves FG, HD and JE are concurrent at K.
QED