Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to enlarge the problem 939.

## Tuesday, December 10, 2013

### Geometry Problem 939: Circle, Tangent, Secant, Chord, Parallel, Midpoint

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<EDC=<ECA (tangent angle theorem) and <EDC=<EAF (parallel). That makes circumcircle of EAC tangent to FA and then by powerpoint theorem, FA^2=FE*FC. But FB^2=FE*FC, so FB^2=FA^2

ReplyDeleteConstruct circle circumscribe AEC

ReplyDeleteAngleBAE=angleEDC=angleECA

So AB is tangent to circle O and also to new constructed circle.

Further as FEC is the common chord of both circles it will bisect the tangent segment

QED

Triangles AEF,DEC & CAF are similar (Apply alternate segment theorem)

ReplyDeleteConsidering AEF & CAF

=> AF/FE = CF/FA

=> AF^2 = FC.FE = BF^2

FB is tangent to circle BCE

ReplyDeleteFA is tangent to circle ACE

Follows that AF = BF

Sumith Peiris

Moratuwa

Sri Lanka