Saturday, February 27, 2016

Geometry Problem 1191: Circle, Chords, Diameter, Congruence, Midpoint, Collinearity, Bisector

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

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Math: Geometry Problem 1191: Circle, Chords, Diameter, Congruence, Midpoint, Collinearity, Bisector
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2 comments:

  1. Peter TranFebruary 27, 2016 at 10:31 PM

    1. Consider hexagon EGDCHF
    we have P= EF∩DC, O=HC∩EG,J=DG∩HF
    so P, O, J are collinear per Pascal’s theorem
    2. In circle O , since PA=PB => JP⊥ AB
    3. Observe that FJPK and JPDL are cyclic quadrilateral
    So ∡ (KJP)= ∡ (KFP)= ∡ (EDC)= ∡ (PJL)
    So JO bisect ∡ (KJL)
    4. KJL is isosceles tri. => JK=JL and KA=BL
    5. Since KO=LO => power of K or L to circle O have the same value= KC.KF= LE.LD

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    Replies
    1. Sumith PeirisFebruary 28, 2016 at 4:37 PM

      Good work Peter in noting to apply Pascal to the cyclic hexagon and showing that J,O,P are collinear after which the rest is fairly straightforward.

      On the last point, KF.KC = KA.KB = BL.LA (since KA = BL) = LD.LE

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