:400px Let the points be labelled to simplify: :$A := \tuple {1, 4}$ :$B := \tuple {2, 8}$ :$C := \tuple {4, 2}$ :$D := \tuple {8, 5}$ :$E := \tuple {7, 1}$ :$F := \tuple {5, 7}$ Let $ABCDEF$ be considered as a hexagon. We join the opposite points of $ABCDEF$: :$AF: \tuple {1, 4} \to \tuple {5, 7}$ :$BC: \tuple {2, 8} \to \tuple {4, 2}$ :$BE: \tuple {2, 8} \to \tuple {7, 1}$ :$AD: \tuple {1, 4} \to \tuple {8, 5}$ :$CD: \tuple {4, 2} \to \tuple {8, 5}$ :$EF: \tuple {7, 1} \to \tuple {5, 7}$ It is to be shown that the intersections of: :$AF$ and $BC$ :$BE$ and $AD$ :$CD$ and $EF$ all lie on the same straight line. The result then follows from Pascal's Mystic Hexagram. From Equation of Straight Line in Plane through Two Points: :$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$ Thus: {{begin-eqn}} {{eqn | n = AF       | l = \frac {y - 4} {x - 1}       | r = \frac {7 - 4} {5 - 1}       | c =  }} {{eqn | r = \frac 3 4       | c =  }} {{eqn | ll= \leadsto       | l = 4 \paren {y - 4}       | r = 3 \paren {x - 1}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac 3 4 x + \dfrac {13} 4       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = BC       | l = \frac {y - 8} {x - 2}       | r = \frac {2 - 8} {4 - 2}       | c =  }} {{eqn | r = -3       | c =  }} {{eqn | ll= \leadsto       | l = y - 8       | r = -3 \paren {x - 2}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = -3 x + 14       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = BE       | l = \frac {y - 8} {x - 2}       | r = \frac {1 - 8} {7 - 2}       | c =  }} {{eqn | r = -\frac 7 5       | c =  }} {{eqn | ll= \leadsto       | l = 5 \paren {y - 8}       | r = -7 \paren {x - 2}       | c =  }} {{eqn | ll= \leadsto       | l = 5 y - 40       | r = -7 x + 14       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = -\frac 7 5 x + \frac {54} 5       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = AD       | l = \frac {y - 4} {x - 1}       | r = \frac {5 - 4} {8 - 1}       | c =  }} {{eqn | r = \frac 1 7       | c =  }} {{eqn | ll= \leadsto       | l = 7 \paren {y - 4}       | r = x - 1       | c =  }} {{eqn | ll= \leadsto       | l = 7 y - 28       | r = x - 1       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \frac x 7 + \frac {27} 7       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = CD       | l = \frac {y - 2} {x - 4}       | r = \frac {5 - 2} {8 - 4}       | c =  }} {{eqn | r = \frac 3 4       | c =  }} {{eqn | ll= \leadsto       | l = 4 \paren {y - 2}       | r = 3 \paren {x - 4}       | c =  }} {{eqn | ll= \leadsto       | l = 4 y - 8       | r = 3 x - 12       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \frac 3 4 x - 1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = EF       | l = \frac {y - 1} {x - 7}       | r = \frac {7 - 1} {5 - 7}       | c =  }} {{eqn | r = -3       | c =  }} {{eqn | ll= \leadsto       | l = y - 1       | r = -3 x + 21       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = -3 x + 22       | c =  }} {{end-eqn}} Evaluate the intersection of $AF$ and $BC$: {{begin-eqn}} {{eqn | l = y       | r = \dfrac 3 4 x + \dfrac {13} 4       | c =  }} {{eqn | l = y       | r = -3 x + 14       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 3 4 x + \dfrac {13} 4       | r = -3 x + 14       | c =  }} {{eqn | ll= \leadsto       | l = 3 x + 13       | r = -12 x + 56       | c =  }} {{eqn | ll= \leadsto       | l = 15 x       | r = 43       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \dfrac {43} {15}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = -3 \paren {\dfrac {43} {15} } + 14       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {-129 + 210} {15}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {81} {15}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {27} 5       | c =  }} {{end-eqn}} So $AF$ and $BC$ intersect at $\paren {\dfrac {43} {15}, \dfrac {27} 5}$. Evaluate the intersection of $BE$ and $AD$: {{begin-eqn}} {{eqn | l = y       | r = -\frac 7 5 x + \frac {54} 5       | c =  }} {{eqn | l = y       | r = \frac x 7 + \frac {27} 7       | c =  }} {{eqn | ll= \leadsto       | l = -\frac 7 5 x + \frac {54} 5       | r = \frac x 7 + \frac {27} 7       | c =  }} {{eqn | ll= \leadsto       | l = -49 x + 378       | r = 5 x + 135       | c =  }} {{eqn | ll= \leadsto       | l = 54 x       | r = 243       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \dfrac 9 2       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \frac 1 7 \paren {\frac 9 2} + \frac {27} 7       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {9 + 54} {14}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {63} {14}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac 9 2       | c =  }} {{end-eqn}} So $BE$ and $AD$ intersect at $\paren {\dfrac 9 2, \dfrac 9 2}$. Evaluate the intersection of $CD$ and $EF$: {{begin-eqn}} {{eqn | l = y       | r = \frac 3 4 x - 1       | c =  }} {{eqn | l = y       | r = -3 x + 22       | c =  }} {{eqn | ll= \leadsto       | l = \frac 3 4 x - 1       | r = -3 x + 22       | c =  }} {{eqn | ll= \leadsto       | l = \frac 3 x - 4       | r = -12 x + 88       | c =  }} {{eqn | ll= \leadsto       | l = 15 x       | r = 92       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \dfrac {92} {15}       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = -3 \paren {\dfrac {92} {15} } + 22       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {-92 + 110} 5       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {18} 5       | c =  }} {{end-eqn}} So $CD$ and $EF$ intersect at $\paren {\dfrac {92} {15}, \dfrac {18} 5}$. It remains to be shown that those points of intersection: :$\paren {\dfrac {43} {15}, \dfrac {27} 5}$, $\paren {\dfrac 9 2, \dfrac 9 2}$, $\paren {\dfrac {92} {15}, \dfrac {18} 5}$ all lie on the same straight line. From Equation of Straight Line in Plane through Two Points: :$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$ Thus: {{begin-eqn}} {{eqn | l = \frac {y - \frac {27} 5} {x - \frac {43} {15} }       | r = \frac {\frac {18} 5 - \frac {27} 5} {\frac {92} {15} - \frac {43} {15} }       | c =  }} {{eqn | ll= \leadsto       | l = \frac {5 y - 27} {15 x - 43}       | r = \frac {18 - 27} {92 - 43}       | c = simplifying  }} {{eqn | ll= \leadsto       | l = \frac {5 y - 27} {15 x - 43}       | r = -\frac 9 {49}       | c =  }} {{eqn | ll= \leadsto       | l = 245 y - 1323       | r = - 135 x + 387       | c =  }} {{eqn | ll= \leadsto       | l = 245 y       | r = - 135 x + 1710       | c =  }} {{eqn | ll= \leadsto       | l = 49 y + 27 x       | r = 342       | c =  }} {{end-eqn}} It remains to demonstrate that $\paren {\dfrac 9 2, \dfrac 9 2}$ lies on this line: {{begin-eqn}} {{eqn | l = 49 \dfrac 9 2 + 27 \dfrac 9 2       | r = \dfrac {441 + 243} 2       | c =  }} {{eqn | r = \dfrac {684} 2       | c =  }} {{eqn | r = 342       | c =  }} {{end-eqn}} Bingo. {{qed}}