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.

Click the figure below to view more details of problem 1191.

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

2 comments:

  1. 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. 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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