# Difference between revisions of "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 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, $\angle CAB \equiv \angle MAB \equiv \angle MNB \equiv \angle MND \equiv \angle MCD \equiv \angle ACD$.

This implies that $AB$ and $CD$ are parallel.

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, $A_1P_3 = A_1A_6$ is parallel to $B_1P_2$, and $A_1A_4$ is parallel to $B_1B_2$, and $P_3A_4 = A_4A_3$ is parallel to $P_2B_2$. It follows that triangles $P_3A_1A_4$ and $P_2B_1B_2$ are homothetic, so the line $P_3P_2$ passes through the intersection of lines $A_1B_1$ (which is the same as line $A_1A_2$) and $A_4B_2$ (which is the same as line $A_4A_5$), which interesect at $P_1$.