Classical Theorem

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## Monday, May 9, 2011

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## Monday, May 9, 2011

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Geometry Problem 826: Brianchon's Theorem, Circumscribed Hexagon, Concurrency

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Online Geometry theorems, problems, solutions, and related topics.

Labels:
Brianchon theorem,
circumscribed,
concurrent,
hexagon

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Geometry Problem 826: Brianchon Theorem, Circumscribed Hexagon, Concurrency

ReplyDeleteCase 1 : If EF // HG and HE // GF, it is trivial.

ReplyDeleteCase 2 : WLOG, assume EF intersects HG. Let EG and FH meet at X. We will prove that BD also passes through X. Let P be the intersection of EF and GH. Then the polar of P (with respect to the given circle) passes through X.

On the other hand, EF, the polar of B, passes through B. So the polar of P passes through B & D.

Since the polar of P passes through B, X and D, they are collinear.

Q.E.D.

Use the same notation as Problem 827,

ReplyDeletelet the tangency points be G,H,J,K,L,M.

Then

GM is the polar of A,

GH is the polar of B,

HJ is the polar of C,

JK is the polar of D,

KL is the polar of E,

LM is the polar of F.

Thus, GHJKLM is a cyclic hexagon.

By Pascal's theorem, (see below)

the points of intersection of the opposite sides,

namely, GH∩KL, HJ∩LM and JK∩GM, are collinear.

Since the intersections of polars are collinear,

by duality, the line joining the poles are concurrent.

Since the line joining the poles are AD, BE and CF,

hence, AD, BE, CF are concurrent.

Pascal's theorem:

http://gogeometry.com/circle/pascal_theorem_proof.htm

here is a video-proof for both Pascal and Brianchon's theorem (into spanish, sorry)

ReplyDeletehttp://www.youtube.com/watch?v=yZ6WhoSTXGA&feature=youtu.be