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README

************************************************************
  README file, for inputs to prover
  CORE Library version 1.2
  $Id: README,v 1.1 2006/03/07 04:54:38 exact Exp $
************************************************************

Simson's Theorem:

        Let D be a point on the circumscribed circle
        (0) of triangle ABC.  From D three perpendiculars
        are drawn to the three sides BC, CA and AB.
        Let E, F and G be the three feet, respectively.
        Show that E, F and G are collinear.

        Setup:  A(0,0), B(u_1,0), C(u_2, u_3), O(x_2, x_1),
        D(x_3, u_4), E(x_5, x_4), F(x_7, x_6), G(x_3, 0)
        Therefore the hypothesis equations are:

        h_1 = 2 u_2 x_2 + 2 u_3 x_1 - u_3^2 - u_2^2 = 0
	h_2 = 2 u_1 x_2 - u_1^2 = 0
	h_3 = -x_3^2 + 2 x_2 x_3 + 2 u_4 x_1 - u_4^2 = 0
	h_4 = u_3 x_5 - u_2 x_4 + u_1 x_4 - u_1 u_3 = 0
	h_5 = u_2 x_5 - u_1 x_5 + u_3 x_4 - u_2 x_3 + u_1 x_3 - u_3 u_4 = 0
	h_6 = u_3 x_7 - u_2 x_6 = 0
	h_7 = u_2 x_7 + u_3 x_6 - u_2 x_3 - u_3 u_4 = 0

	The conclusion that points E,F and G are collinear 
	is equivalent to:

	x_4 x_7 - x_5 x_6 + x_3 x_6 - x_3 x_4 = 0	


Pascal's Theorem:

	Let A, B, C, D, F and E be six points on a circle 
	(O). Let P intersection of AB and DF, Q intersection 
	of BC and FE and S intersection of CD and EA.
	Show that P, Q and S are collinear.

	Let A(0,0), O(u_1,0), B(x_1,u_2), C(x_2,u_3), D(x_3,u_4)
	F(x_4,u_5), E(x_5,u_6), P(x_7,x_6), Q(x_9,x_8) and S(x_11,x_10).
	Therefore the hypothesis equations are:

	h_1 = x_1^2 - 2 u_1 x_1 + u_1^2 = 0 	
	h_2 = x_2^2 - 2 u_1 x_2 + u_3^2 = 0 
	h_3 = x_3^2 - 2 u_1 x_3 + u_4^2 = 0 
	h_4 = x_4^2 - 2 u_1 x_4 + u_5^2 = 0 
	h_5 = x_5^2 - 2 u_1 x_5 + u_6^2 = 0 
	h_6 = u_5 x_7 - u_4 x_7 - x_4 x_6 + x_3 x_6 + u_4 x_4 - u_5 x_3 = 0
	h_7 = u_2 x_7 - x_1 x_6 = 0 
	h_8 = u_6 x_9 - u_5 x_9 - x_5 x_8 + x_4 x_8 + u_5 x_5 - u_6 x_4 = 0 
	h_9 = u_3 x_9 - u_2 x_9 - x_2 x_8 + x_1 x_8 + u_2 x_2 - u_3 x_1 = 0 
	h_10 = u_6 x_11 - x_5 x_10 = 0 
	h_11 = u_4 x_11 - u_3 x_11 - x_3 x_10 + x_2 x_10 + u_3 x_3 - u_4 x_2 = 0 

	The conclusion that P, Q and S are collinear is:

	x_8 x_11 - x_6 x_11 - x_9 x_10 + x_7 x_10 + x_6 x_9 - x_7 x_8 = 0


Butterfly Theorem:

	Let A, B, C, D be four points on a circle (O).
	E is the intersection of AC and BD. Through E 
	draw a line perpendicular to OE, meeting AD at 
	F and BC at G. Show that FE=GE.

	Let E(0,0), O(u_1,0), A(u_2,u_3), B(x_1,u_4),
	C(x_3,x_2), D(x_5,x_4), F(0,x_6) and G(0,x_7).

	Therefore the hypothesis equations are:

	h_1 = x_1^2 - 2 u_1 x_0 + u_3^2 - u_3^2 - u_2^2 + 2u_1 u_2 = 0 
	h_2 = x_3^2 - 2u_1 x_3 + x_2^2 - u_3^2 - u_2^2 + 2u_1 u_2 = 0 
	h_3 = -u_3 x_3 + u_2 x_2 = 0
	h_4 = x_5^2 - 2u_1 x_5 + x_4^2 - u_3^2 - u_2^2 + 2u_1 u_2 = 0 
	h_5 = -u_4 x_5 + x_1 x_4 = 0
	h_6 = -x_5 x_6 + u_2 x_6 + u_3 x_5 - u_2 x_4 = 0 
	h_7 = -x_3 x_7 + x_1 x_7 + u_4 x_3 - x_1 x_2 = 0

	The conclusion is:

	g = x_7 + x_6 = 0


Pappus's Theorem:

	Let ABC and EFG be two lines and P be the intersection
	of AF and EB, Q the intersection of AG and EC and S the
	intersectin of BG and FC. Then P, Q and S are collnear.

	Let A(0,0), B(u1,0), C(u_2,0), E(u_2,u_3), F(u_4,u_5),
	G(x_1,u_7), P(x_3,x_2), Q(x_5,x_4), S(x_7,x_6).

	Therefore the hypothesis equations are: 

	h_1 = u_5 x_1 - u_3 x_1 - u_2 u_5 - u_4 u_7 + 
	      u_3 u_4 + u_2 u_7 = 0
	h_2 = u_4 x_2 - u_5 x_3 = 0
	h_3 = u_2 x_2 - u_1 x_2 - u_3 x_3 + u_3 u_1 = 0
	h_4 = x_1 x_4 - u_7 x_5 = 0
	h_5 = u_2 x_4 - u_6 x_4 - u_3 x_5 + u_3 u_6 = 0
	h_6 = u_5 x_6 - u_6 x_6 - u_5 x_7 + u_5 u_6 = 0
	h_7 = x_1 x_6 - u_1 x_6 - u_7 x_7 + u_1 u_7 = 0

	The conclusion is:

	g = x_7 x_4 - x_2 x_7 - x_3 x_4 - x_5 x_6 +
	    x_3 x_6 + x_2 x_5 - x_2 x_5 = 0

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