[alogo] 1. An application of Pappus theorem

Let ABC be a triangle and (X,Y), (X',Y') be respectively pairs of points on the sides AB, AC. Then the intersection points  of lines :  H=(BX',CX), I=(BY',CY) and J=(XY',XY') are collinear.
This is a direct application of Pappus theorem (see PappusLines.html ).

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[alogo] 2. Application

Let ABC be a triangle and (A',B',C') be the middles of the sides and (A'',B'',C'') the feet of the altitudes.
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.

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[alogo] 3. Euler as Pascal

The Euler line of the triangle ABC is the pascal line of the hexagon C'B''A'C''B'A'', which is inscribed in the Euler circle (see Pascal.html ). This is a direct corrolary of the previous application.

[alogo] 4. An interesting figure

Next figure is created by drawing the diagonals of hexagons inscribed in the Euler circle and having vertices the points A',B',C',A'',B'',C''. The figure suggests some remarkable coincidences.

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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'').

Property-1 Points {A**,C**,B*} are collinear. Analogously {B**,A**,C*} and {C**,B**,A*}.
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.

Property-2 The polar p(A*) = AA**. Analogously p(B*)=BB**, p(C*)=CC**. These three polars are orthogonal to the Euler line, hence parallel.
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.

Property-3  B** is the middle of  BB0, where B0 is the intersection B0=(AC,BB**). Analogous properties for C** and C0 and A** and A0.
In fact, B** is on A'C' which is the medial line of the altitude BB''. A'C' being parallel to AC intersects B0
at its middle.

Property-4 B** is on line AA*. Analogous properties for C** and BB* and A** and CC*.
Since B** is the middle of BB0 and AA** is parallel to BB**, the pencil of lines A(B,C,A**,B**) is a harmonic one. But the pencil A(B,C,A**,A*) is also harmonic, hence AA* and AB** coincide.

Corrolary Triangle A**B**C** is autopolar w.r. to the Euler circle.

See Also

Euler line and circle
Harmonic Pencils
Polar Construction

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