Online Geometry theorems, problems, solutions, and related topics.
Classical TheoremClick the figure below to see the complete classical theorem.
Geometry Problem 827: Brianchon Corollary, Circumscribed Hexagon, Concurrency lines.
let GM,JK intersect at X. Say GK and JM intersects at N. N lies on the polar of X.Also, A is the pole of polar GM and D is the pole of polar JK.By LaHire Theorem, AD is the polar of the intersection of GM,JK (i.e. Point X)Therefore, A,N,D are collinear.Q.E.D.
To W FungRefer to line 2 "Say GK and JM intersects at N. N lies on the polar of X."I am not sure how you get this. If this is the case, we can conclude that A,N, D are collinear since AD is the polar of intersecting point of GM and JK. Please explain.
This would be a common properties for Pole and Polar application. If you insist, you may refer to Brokard's Theorem.
what is meant by polar
The polar line of a point is a line perpendicular to line joining the point and the center of the circle, and it must contain the inverse of the point.
http://img844.imageshack.us/img844/6590/problem827.pngLet DE cut AF at P and AB cut DC at QLet N is the intersecting point of AD and PQ1. Staring from hexagon ABCDEF ,Let B approach Q then H and C coincide to JThen Hexagon ABCDEF become pentagon AQDEF- Diagonal AD will stay the same - Diagonal EB become EQ- Diagonal FC become FJ Per Brianchon’s theorem AD, EQ and JF are concurrent at point R ( see sketch)2. Starting from pentagon AQDEF , let E approach P then L and F coincide to M Then pentagon AQDEF become quadrilateral AQDP- Diagonal QE become QP - JF become JM- EQ become PQPoint of concurrent R will become N => PQ, AD and JM will concurrent at N3. From hexagon ABCDEF , Let C approach Q then B and H coincide to G Then hexagon become pentagon AQDEFSimilar to step 1, QF, GE and AD will concurrent at point S ( not shown)4. From pentagon AQDEF, let F approach P then pentagon become quadrilateral QAPD andL and E coincide to KSimilar to step 2 , point of concurrent S will become NSo PQ, AD, GK and JM will concurrent at point N
Let AB and CD meet at P, AF and DE meet at Q. Then AHDQ is a tangential quadrilateral. By a theorem, AD,GK,MJ,PQ are concurrent. i.e. AD,GK,MJ are concurrent at N. Remark. The theorem I used can be view as a corollary of Brianchon theorem, but it can be proved by other arguments.