This is a direct application of Pappus theorem (see PappusLines.html ).

Then the three intersection points of lines A*=(C'B'',B'C''), B*=(A'C'',C'A''), C*=(B'A'',A'B''), are located on the Euler line.

In the figure the new points are intersections of lines:

A**=(B'C',B''C''), B**=(C'A',C''A'') and C**=(A'B',A''B'').

This follows from the fact that the three points are intersections of opposite sides of the hexagon C'B'A'C''B''A'',

inscribed in the Euler circle. Thus, by Pascal's theorem, they are collinear.

This follows from Nr. 2 above. In fact the polar p(A*) passes through A and A**, since these points are diagonal points of the quadrilateral C'C''B'B'', which is inscribed in the Euler circle. Besides A* being on the Euler line,

p(A*) is orthogonal to the Euler circle.

In fact, B** is on A'C' which is the medial line of the altitude BB''. A'C' being parallel to AC intersects B

at its middle.

Since B** is the middle of BB

Harmonic Pencils

PappusLines.html

Pascal.html

Polar Construction

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