Theorem([arbitrary(C,A,G), oncircle(A,C,G,B), perpfoot(A,C,G,H,H), perpfoot(A,C,B,E,E), perpfoot(B,C,A,D,D), parallel(D,A,J,G), online(B,I,A), intersection(B,E,D,A,F), perpfoot(B,C,G,J,J), perpendicular(B,I,I,G), parallel(B,E,H,G), midpoint(F,G,M), online(I,J,H)],[online(I,J,M)]) Warning: the coordinates of the points from {I, A, B, C, D, E, F, G, H, J, M} are reassigned Theorem: If the points C, A, and G are arbitrary, the point B is on the circumcircle of the triangle ACG, H is the perpendicular foot of the line AC to the line GH, E is the perpendicular foot of the line AC to the line BE, D is the perpendicular foot of the line BC to the line AD, the line DA is parallel to the line JG, `the point A is on the line B` || (I) , the two lines BE and DA intersect at F, J is the perpendicular foot of the line BC to the line GJ, `the line B` || (I) || ` is perpendicular to the line ` || (I) || G , the line BE is parallel to the line HG, M is the midpoint of F and G, and `the point H is on the line ` || (I) || J , then `the point M is on the line ` || (I) || J . Proof: Status: used = 770, alloced = 3072, time = 4.601999999999999 char set produced: [7 y1 2] [1 x2 1] [1 y2 1] [2 x3 1] [1 y3 1] [8 x4 1] [4 y4 1] [7 x5 1] [4 y5 1] [12 x6 1] [4 y6 1] [12 x7 1] [6 y7 1] [2 x8 1] [3 y8 1] char set recomputed: [7 y1 2] [1 x2 1] [1 y2 1] [2 x3 1] [1 y3 1] [12 x4 1] [4 y4 1] [8 x5 1] [4 y5 1] [25 x6 1] [4 y6 1] [2 x7 1] [10 y7 1] [2 x8 1] [3 y8 1] pseudo-remainder: [6 y8 1] = x5*y8 + ... (5 terms) pseudo-remainder: [8 x8 1] = -2*y5*x8 + ... (7 terms) pseudo-remainder: [8 y7 1] = y7*x5 + ... (7 terms) pseudo-remainder: [40 x7 1] = u1*u2*v1*x7*y5 + ... (39 terms) pseudo-remainder: [40 y6 1] = x1*u1*u2*v1*y6 + ... (39 terms) pseudo-remainder: [59 x6 1] = v1^2*u2^2*u1*x6 + ... (58 terms) pseudo-remainder: [246 y5 1] = 2*u1^4*u2^4*v1^2*y5 + ... (245 terms) pseudo-remainder: [186 x5 1] = u1^5*u2^2*v1^3*x5 + ... (185 terms) pseudo-remainder: [342 x4 1] = u1^6*x4*v1^4*u2^2 + ... (341 terms) pseudo-remainder: [425 y1 5] = -2*u1^7*y1^5*u2^2*v1^2 + ... (424 terms) `The theorem is true under the following subsidiary conditions:` . the three points A, C and G are not collinear . the line CG is non-isotropic . the line AG is not perpendicular to the line AC . v1*u1^2-u1*u2*v1-u1*v1*x1+u1*y1*u2+y1*v1^2+x1*u2*v1 <> 0 . -u1*v1*x1+u1*u2*v1-u1*y1*u2-x1*u2*v1-y1*v1^2+v1*x1^2 <> 0 . -u1^2*v1*x1*u2+u1^2*v1*u2^2-u1^2*y1*u2^2+x1^2*v1*u2*u1-u1*v1^2*x1*y1-y1*v1^2*u2*u1-u1*x1*v1^3+u1*v1^3*u2-x1*v1*u2^2*u1+x1^2*v1^3-u2*v1^2*x1*y1+y1*v1^2*x1^2-y1*v1^4-u2*x1*v1^3 <> 0 . the three points A, B and C are not collinear . the line AC is not perpendicular to the line AB . the line AC is not perpendicular to the line BC QED.
Theorem([arbitrary(C,A,G), oncircle(A,C,G,B), perpfoot(A,C,G,H,H), perpfoot(A,C,B,E,E), perpfoot(B,C,A,D,D), parallel(D,A,J,G), online(B,I,A), intersection(B,E,D,A,F), perpfoot(B,C,G,J,J), perpendicular(B,I,I,G), parallel(B,E,H,G), midpoint(F,G,M), online(I,J,M)],[online(I,J,H)]) Warning: the coordinates of the points from {I, A, B, C, D, E, F, G, H, J, M} are reassigned Theorem: If the points C, A, and G are arbitrary, the point B is on the circumcircle of the triangle ACG, H is the perpendicular foot of the line AC to the line GH, E is the perpendicular foot of the line AC to the line BE, D is the perpendicular foot of the line BC to the line AD, the line DA is parallel to the line JG, `the point A is on the line B` || (I) , the two lines BE and DA intersect at F, J is the perpendicular foot of the line BC to the line GJ, `the line B` || (I) || ` is perpendicular to the line ` || (I) || G , the line BE is parallel to the line HG, M is the midpoint of F and G, and `the point M is on the line ` || (I) || J , then `the point H is on the line ` || (I) || J . Proof: Status: used = 3232, alloced = 5506, time = 4.758000000000001 Status: used = 12372, alloced = 13080, time = 5.6000000000000005 char set produced: [7 y1 2] [1 x2 1] [1 y2 1] [2 x3 1] [1 y3 1] [8 x4 1] [4 y4 1] [7 x5 1] [4 y5 1] [81 x6 1] [4 y6 1] [12 x7 1] [6 y7 1] [2 x8 1] [3 y8 1] char set recomputed: [7 y1 2] [1 x2 1] [1 y2 1] [2 x3 1] [1 y3 1] [12 x4 1] [4 y4 1] [8 x5 1] [4 y5 1] [473 x6 1] [4 y6 1] [2 x7 1] [10 y7 1] [2 x8 1] [3 y8 1] pseudo-remainder: [6 y6 1] = y6*x2 + ... (5 terms) pseudo-remainder: [12 x6 1] = u2*x6*y2 + ... (11 terms) pseudo-remainder: [1784 y5 1] = 2*u1^9*u2^9*v1*y5 + ... (1783 terms) pseudo-remainder: [1571 x5 1] = 2*u1^10*u2^7*v1*x5*y2 + ... (1570 terms) pseudo-remainder: [3048 x4 1] = 2*u1^11*u2^7*v1^2*x4*y2 + ... (3047 terms) pseudo-remainder: [3549 y2 1] = 2*u1^13*u2^8*v1^3*y2 + ... (3548 terms) pseudo-remainder: [1444 x2 1] = -2*u1^11*u2^7*v1^2*x2*y1^2 + ... (1443 terms) pseudo-remainder: [695 y1 3] = 2*y1^3*v1*u1^10*u2^7 + ... (694 terms) `The theorem is true under the following subsidiary conditions:` . the three points A, C and G are not collinear . the line AG is not perpendicular to the line AC . v1*u1^2-u1*u2*v1-u1*v1*x1+u1*y1*u2+y1*v1^2+x1*u2*v1 <> 0 . -u1*v1*x1+u1*u2*v1-u1*y1*u2-x1*u2*v1-y1*v1^2+v1*x1^2 <> 0 . -u1^2*v1*x1*u2+u1^2*v1*u2^2-u1^2*y1*u2^2+x1^2*v1*u2*u1-u1*v1^2*x1*y1-y1*v1^2*u2*u1-u1*x1*v1^3+u1*v1^3*u2-x1*v1*u2^2*u1+x1^2*v1^3-u2*v1^2*x1*y1+y1*v1^2*x1^2-y1*v1^4-u2*x1*v1^3 <> 0 . 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+20*u1*v1^12*x1^3*y1-12*u1*v1^10*x1^5*y1-9*u1^7*u2^2*v1^5*x1^3-38*u1^5*u2^5*v1^5*x1^2+3*u1^2*u2^2*v1^6*x1^6*y1-20*u1^2*u2*v1^12*x1*y1-u1^2*u2*v1^6*x1^7*y1-14*u1*u2^2*v1^12*x1*y1-12*u1*u2*v1^10*x1^4*y1+8*u1*u2*v1^8*x1^6*y1 <> 0 . the three points A, B and C are not collinear . the line AC is not perpendicular to the line AB . the line AC is not perpendicular to the line BC QED.
Theorem([arbitrary(C,G,A), oncircle(A,C,G,B), perpfoot(A,C,G,H,H), perpfoot(A,C,B,E,E), perpfoot(B,C,A,D,D), parallel(D,A,J,G), online(B,I,A), intersection(B,E,D,A,F), perpfoot(B,C,G,J,J), perpendicular(B,I,I,G), midpoint(F,G,M), online(I,J,H), online(I,J,M)],[parallel(B,E,H,G)]) Warning: the coordinates of the points from {I, A, B, C, D, E, F, G, H, J, M} are reassigned Theorem: If the points C, G, and A are arbitrary, the point B is on the circumcircle of the triangle ACG, H is the perpendicular foot of the line AC to the line GH, E is the perpendicular foot of the line AC to the line BE, D is the perpendicular foot of the line BC to the line AD, the line DA is parallel to the line JG, `the point A is on the line B` || (I) , the two lines BE and DA intersect at F, J is the perpendicular foot of the line BC to the line GJ, `the line B` || (I) || ` is perpendicular to the line ` || (I) || G , M is the midpoint of F and G, `the point H is on the line ` || (I) || J , and `the point M is on the line ` || (I) || J , then the line BE is parallel to the line HG. Proof: Status: used = 31806, alloced = 23808, time = 7.862 char set produced: [7 y1 2] [4 x2 1] [3 y2 1] [5 x3 1] [3 y3 1] [7 x4 1] [4 y4 1] [8 x5 1] [4 y5 1] [10 x6 1] [4 y6 1] [10 x7 1] [6 y7 1] [3 x8 1] [2 y8 1] pseudo-remainder: [6 y3 1] = -u2*y3 + ... (5 terms) pseudo-remainder: [8 x3 1] = -u1*x3*y2 + ... (7 terms) pseudo-remainder: [9 y2 1] = u1^2*v1^2*y2 + ... (8 terms) pseudo-remainder: [12 x2 1] = -u1^3*v1*x2 + ... (11 terms) `The theorem is true under the following subsidiary conditions:` . the three points A, C and G are not collinear . the line AC is non-isotropic . the line BC is non-isotropic . the line BE is not parallel to the line DA . the line AC is not perpendicular to the line CG . the line AB is not perpendicular to the line CG . the line BC is not perpendicular to the line CG . the line BE is not perpendicular to the line CG . the line AB is not parallel to the line HJ QED.
Theorem([arbitrary(C,G,A), oncircle(A,C,G,B), perpfoot(A,C,G,H,H), perpfoot(A,C,B,E,E), perpfoot(B,C,A,D,D), parallel(D,A,J,G), online(B,I,A), intersection(B,E,D,A,F), perpfoot(B,C,G,J,J), parallel(B,E,H,G), midpoint(F,G,M), online(I,J,H), online(I,J,M)],[perpendicular(B,I,I,G)]) Warning: the coordinates of the points from {I, A, B, C, D, E, F, G, H, J, M} are reassigned Theorem: If the points C, G, and A are arbitrary, the point B is on the circumcircle of the triangle ACG, H is the perpendicular foot of the line AC to the line GH, E is the perpendicular foot of the line AC to the line BE, D is the perpendicular foot of the line BC to the line AD, the line DA is parallel to the line JG, `the point A is on the line B` || (I) , the two lines BE and DA intersect at F, J is the perpendicular foot of the line BC to the line GJ, the line BE is parallel to the line HG, M is the midpoint of F and G, `the point H is on the line ` || (I) || J , and `the point M is on the line ` || (I) || J , then `the line B` || (I) || ` is perpendicular to the line ` || (I) || G . Proof: char set produced: [7 y1 2] [4 x2 1] [3 y2 1] [5 x3 1] [3 y3 1] [7 x4 1] [4 y4 1] [8 x5 1] [4 y5 1] [10 x6 1] [4 y6 1] [10 x7 1] [6 y7 1] [3 x8 1] [2 y8 1] pseudo-remainder: [6 y6 2] = -y6^2 + ... (5 terms) pseudo-remainder: [12 x6 2] = -v1^2*x6^2 + ... (11 terms) pseudo-remainder: [98 y5 2] = -x1^5*y5^2*u2 + ... (97 terms) pseudo-remainder: [212 x5 2] = u1^2*x1*v1*x5^2*y1*u2 + ... (211 terms) pseudo-remainder: [700 y2 2] = -u1^6*u2^2*y2^2*v1^2 + ... (699 terms) pseudo-remainder: [440 x2 2] = -v1^2*x2^2*u1^5*u2^2*x1 + ... (439 terms) pseudo-remainder: [244 y1 7] = u1^8*v1*y1^7 + ... (243 terms) `The theorem is true under the following subsidiary conditions:` . the three points A, C and G are not collinear . the line AC is non-isotropic . the line BC is non-isotropic . the line BE is not parallel to the line DA . the line AC is not perpendicular to the line CG . the line AB is not perpendicular to the line CG . the line BC is not perpendicular to the line CG . the line BE is not perpendicular to the line CG . the line AB is not parallel to the line HJ QED.
Theorem([arbitrary(C,A,G), oncircle(A,C,G,B), perpfoot(A,C,G,H,H), perpfoot(A,C,B,E,E), perpfoot(B,C,A,D,D), online(B,I,A), intersection(B,E,D,A,F), perpfoot(B,C,G,J,J), perpendicular(B,I,I,G), parallel(B,E,H,G), midpoint(F,G,M), online(I,J,H), online(I,J,M)],[parallel(D,A,J,G)]) Warning: the coordinates of the points from {I, A, B, C, D, E, F, G, H, J, M} are reassigned Theorem: If the points C, A, and G are arbitrary, the point B is on the circumcircle of the triangle ACG, H is the perpendicular foot of the line AC to the line GH, E is the perpendicular foot of the line AC to the line BE, D is the perpendicular foot of the line BC to the line AD, `the point A is on the line B` || (I) , the two lines BE and DA intersect at F, J is the perpendicular foot of the line BC to the line GJ, `the line B` || (I) || ` is perpendicular to the line ` || (I) || G , the line BE is parallel to the line HG, M is the midpoint of F and G, `the point H is on the line ` || (I) || J , and `the point M is on the line ` || (I) || J , then the line DA is parallel to the line JG. Proof: char set produced: [7 y1 2] [1 x2 1] [1 y2 1] [2 x3 1] [1 y3 1] [8 x4 1] [4 y4 1] [35 x5 1] [4 y5 1] [12 x6 1] [6 y6 1] [7 x7 1] [4 y7 1] [2 x8 1] [3 y8 1] Status: used = 75549, alloced = 23808, time = 11.590000000000003 char set recomputed: [7 y1 2] [1 x2 1] [1 y2 1] [2 x3 1] [1 y3 1] [12 x4 1] [4 y4 1] [25 x5 1] [4 y5 1] [2 x6 1] [10 y6 1] [12 x7 1] [4 y7 1] [2 x8 1] [3 y8 1] pseudo-remainder: [5 y7 1] = -u2*y7 + ... (4 terms) pseudo-remainder: [10 x7 1] = -u1*y4*x7 + ... (9 terms) pseudo-remainder: [40 y4 1] = u1^3*v1*u2*y4 + ... (39 terms) pseudo-remainder: [32 x4 1] = -v1^2*u1^3*x4 + ... (31 terms) pseudo-remainder: [22 y1 2] = -y1^2*v1^2*u1^2 + ... (21 terms) `The theorem is true under the following subsidiary conditions:` . the three points A, C and G are not collinear . the line CG is non-isotropic . the line AG is not perpendicular to the line AC . v1*u1^2-u1*u2*v1-u1*v1*x1+u1*y1*u2+y1*v1^2+x1*u2*v1 <> 0 . -u1*v1*x1+u1*u2*v1-u1*y1*u2-x1*u2*v1-y1*v1^2+v1*x1^2 <> 0 . -u1^2*v1*x1*u2+u1^2*v1*u2^2-u1^2*y1*u2^2+x1^2*v1*u2*u1-u1*v1^2*x1*y1-y1*v1^2*u2*u1-u1*x1*v1^3+u1*v1^3*u2-x1*v1*u2^2*u1+x1^2*v1^3-u2*v1^2*x1*y1+y1*v1^2*x1^2-y1*v1^4-u2*x1*v1^3 <> 0 . the line AC is not perpendicular to the line AB . the line AC is not perpendicular to the line BC QED.