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

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

Details: Click on the figure below.

## Sunday, November 4, 2018

### Geometry Problem 1397: Circle, Diameter, Tangent, Secant, 90 Degrees Angle, Perpendicular, Collinear Points

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Extend CD to meet AE at P and FB at Q.

ReplyDeletePA & PB are external tangents drawn to the circle as is QD & QB

Hence CQDP bisects BF & AE

But BF//AE,

So it follows from similar triangles that

E,F,C must be collinear

(since if EC meets BF at F’, F’P = PB = FP and so F & F’ coincide)

Sumith Peiris

Moratuwa

Sri Lanka

CD meet BF on Q and AE on P. Tr DQB and DQF isosceles (DQ=QB, angBAF+BFA=90°)

ReplyDelete=> Q midpoint

Tr PAD and PDE isosceles (PA=PD, PE=PD, ang BDC=angPDE,EAD+AED = 90°)

=> P midpoint

From similar tr QB/PA = BC/AC => 2.QB/2.PA = BC/AC => BF/AE = BC/AC