Geometry Problem 1498. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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Since E is on the radical axis of circles [ACG] and [BDF]

ReplyDeleteSo we have ẸC.EG=EB.EF= power of point E to both circles….(1)

Let MN meet AG at H’ and MN meet DF at H

Power of E with circle [H’GKC]= EG.EC= EK.EH’… (2)

Power of E with circle [HFKB] =EB.EF=EK.EH….. (3)

Compare (1),(2) and (3) => EH=EH’ => H coincide to H’

And H, M, N are collinear

this result will help to prove the questions 1 to 6