## A case of inverse/direct Pascal theorem application

Consider a triangle ABC and reflect it on a line a' passing through the middle A' of its side BC to the triangle A*B*C*.
[1] There is a conic ca passing through all six vertices of triangles ABC, A*B*C*.
[2] The parallel to B*C* from A meets ca a second time at a point D lying on the circumcircle c of ABC.

[1] is a consequence of the inverse of Pascal's theorem applied to the hexagon ABB*A*CC*. Its opposite sides intersect at two real points (the side pairs (AB,A*C) and (A*B*,AC*)) and one point at infinity (sides (BB*,CC*)). Hence they are on a line, consequently there is a conic ca passing through all six vertices of the two triangles.
[2] is a consequence of the Pascal theorem. First note that A' is the center of the conic since it is the center of a rectangular paralellogram BB*CC* inscribed in ca. Then define E to be the symmetric of A* with respect to A'. This is again a point of the conic ca. Consider then the hexagon (non-convex) ADBC*B*E and apply there Pascal's theorem. The pairs of opposite sides for this hexagon are (AD,B*C*), (DB, B*E), (BC*,EA). Their intersection points must be collinear. The first and third side pairs are parallel, hence intersect at the line at infinity. The same must be true for the second pair. Hence DB and B*E are parallel. This implies trivially that angle(ADB)=angle(ACB), hence D is the intersection point of ca with the circumcircle.
There is an interesting third property of this figure related to the Euler circle of ABC and examined in the file PascalEuler.html .
Since the symmetric A'' of A with respect to A' is also on the ellipse, the previous remark gives a way to construct all the ellipses circumscribing a given parallelogram ABA''C. Simply take a fifth point D on the circumcircle of ABC and pass a conic through the five points A, B, A'', C and D. EucliDraw has the corresponding tool to do that.