tag:blogger.com,1999:blog-6933544261975483399.post2245158587327678885..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1191: Circle, Chords, Diameter, Congruence, Midpoint, Collinearity, BisectorAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-27657554288528060082016-02-28T16:37:04.692-08:002016-02-28T16:37:04.692-08:00Good work Peter in noting to apply Pascal to the c...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.<br /><br />On the last point, KF.KC = KA.KB = BL.LA (since KA = BL) = LD.LESumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54824011901413508092016-02-27T22:31:24.349-08:002016-02-27T22:31:24.349-08:001. Consider hexagon EGDCHF
we have P= EF∩DC, O=HC...1. Consider hexagon EGDCHF <br />we have P= EF∩DC, O=HC∩EG,J=DG∩HF<br />so P, O, J are collinear per Pascal’s theorem<br />2. In circle O , since PA=PB => JP⊥ AB<br />3. Observe that FJPK and JPDL are cyclic quadrilateral<br />So ∡ (KJP)= ∡ (KFP)= ∡ (EDC)= ∡ (PJL)<br />So JO bisect ∡ (KJL)<br />4. KJL is isosceles tri. => JK=JL and KA=BL<br />5. Since KO=LO => power of K or L to circle O have the same value= KC.KF= LE.LD<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com