Sunday, November 4, 2018

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

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

Posted by Antonio Gutierrez at 12:13 PM
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2 comments:

  1. Sumith PeirisNovember 6, 2018 at 1:06 AM

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

    PA & 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

    ReplyDelete
  2. c.t.e.oNovember 6, 2018 at 11:42 AM

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

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