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

Labels:
90,
angle bisector,
circumcircle,
concurrent,
parallel,
perpendicular,
triangle

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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