[alogo] A remarkable control point on the Euler circle

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.
[3] The intersection point D of the conic ca and c, the centroid G of ABC and the intersection point D*, other than A' of the Euler circle of ABC and B*C* are collinear. Further the ratio DG/GD*=2.
[1] and [2] were discussed in PascalImmortel.html . Here we discuss the third property.

[0_0] [0_1]
[1_0] [1_1]

The third property is a trivial consequence of the considerations in the cited reference. There we proved that AD is parallel to B*C*. Since OA and FA' are also parallel (see Euler.html ) the isoscelli OAD and FA'D* are similar, and the same is true with triangles FGD* and GOD. Hence the intersection point of DD* with OF coincides with the centroid G of ABC and the ratio DG/GD*=OD/FD*=2.
From this property follows that the shape of the conic ca is controlled from the position of point D* on the Euler circle of ABC. One could start from this point, which defines the reflexion axis a' as the bisector of angle BA'B* and point D on line D*G.
Since all conics ca pass also from the symmetric H of A with respect to A' the subject has tight relations with the subject of all conics circumscribing a parallelogram (like ABHC).

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