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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.
1. Consider hexagon EGDCHF we have P= EF∩DC, O=HC∩EG,J=DG∩HFso P, O, J are collinear per Pascal’s theorem2. In circle O , since PA=PB => JP⊥ AB3. Observe that FJPK and JPDL are cyclic quadrilateralSo ∡ (KJP)= ∡ (KFP)= ∡ (EDC)= ∡ (PJL)So JO bisect ∡ (KJL)4. KJL is isosceles tri. => JK=JL and KA=BL5. Since KO=LO => power of K or L to circle O have the same value= KC.KF= LE.LD
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