# Pascal's Theorem $[asy] size(250); D(unitcircle,red); pair A=dir(20), B=dir(40), C=dir(70), D=dir(120), E=dir(210),F=dir(300); D(L(D(IP(D(L(A,B,0.5,7)),D(L(E,D,0.5,1.5)))), D(IP(D(L(C,B,0.5,10)),D(L(E,F,0.5,3.3)))),0.1),dashed); D(IP(D(L(A,F,1,0.5)),D(L(C,D,2,0.5)))); D(A--B--C--D--E--F--A,black+linewidth(1)); [/asy]$ A diagram of the theorem

Pascal's Theorem is a result in projective geometry. It states that if a hexagon is inscribed in a conic section, then the points of intersection of the pairs of its opposite sides are collinear:

Since it is a result in the projective plane, it has a dual, Brianchon's Theorem, which states that the diagonals of a hexagon circumscribed about a conic concur.

## Proof

It is sufficient to prove the result for a hexagon inscribed in a circle, for affine transformations map this circle to any ellipse while preserving collinearity and concurrence in the projective plane, and projective transformations can map an ellipse to any conic while similarly preserving collinearity and concurrence in the projective sense. Thus we will prove the theorem for a cyclic hexagon, using directed angles mod $\pi$.

Lemma. Let $\omega_1, \omega_2$ be two circles which intersect at $M, N$, let $AB$ be a chord of $\omega_1$, and let $C, D$ be the second intersections of lines $AM, BN$ with $\omega_2$. Then $AB$ and $CD$ are parallel.

Proof. Since $ABNM, CDMN$ are two sets of concyclic points and $A,M,C$ and $B,N,D$ are two sets of collinear points, $\angle CAB \equiv \angle MAB \equiv 180^\circ -\angle MNB \equiv \angle MND \equiv 180^\circ - \angle MCD \equiv 180^\circ - \angle ACD$.

Because alternate interior angles $CAB$ and $ACD$ are congruent, $AB || CD.$

Theorem. Let $A_1A_2A_3A_4A_5A_6$ be a cyclic hexagon, and let $P_1 = A_1A_2 \cap A_4A_5$, $P_2 = A_2A_3 \cap A_5A_6$, $P_3 = A_3A_4 \cap A_6A_1$. Then $P_1, P_2, P_3$ are collinear.

Proof. Let $\omega_1$ be the circumcircle of $A_1A_2A_3A_4A_5A_6$, and let $\omega_2$ be the circumcircle of $\triangle{A_2A_5P_2}$. Let $B_1$ be the second intersection of $\omega_2$ with $A_1A_2$, and let $B_2$ be the second intersection of $A_4A_5$ with $\omega_2$. By lemma, $A_1P_3 \parallel B_1P_2$, and $A_1A_4 \parallel B_1B_2$, and $P_3A_4 \parallel P_2B_2$, see figure. It follows that triangles $\triangle{P_3A_1A_4}$ and $\triangle{P_2B_1B_2}$ are homothetic with a center $P_1$, because $A_1B_1 \cap A_4B_2 = P_1$. Therefore the line $P_3P_2$ passes through the center of homothety, i.e. $P_1$.