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## Re: [EMHL] Midcevian Triangle

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• Dear Paul ... The perspectivity can be proven by applying the Fruitful Theorem (nested cevian triangles theorem). Denote T = The reference triangle ABC mT =
Message 1 of 6 , Jan 21, 2004
Dear Paul

[APH]:
>>Definition:
>>
>>Let ABC be a triangle, P a point, A'B'C' the cevian triangle
>>of P and A*,B*,C* the midpoints of the cevians AA', BB', CC' of P.
>>
>>I call the triangle A*B*C* midcevian triangle of P.
>>
>>Locus:
>>
>>Which is the locus of P such that the medial triangle of
>>ABC and the medial triangle of the midcevian triangle of P
>>are perspective? (The entire plane?)

[PY]:
>Yes, this is the point that divides PG in the ratio 3:1.
>
>In fact, the midcevian triangle of P is perspective with the medial
>triangle at the inferior of the isotomic conjugate of P.

The perspectivity can be proven by applying the Fruitful Theorem
(nested cevian triangles theorem).

Denote

T = The reference triangle ABC

mT = The Medial Triangle of ABC

V = The midcevian triangle of P

mV = The medial triangle of the midcevian triangle V of P

We have:

mT is inscribed in T, and T, mT are perspective

V is inscribed in mT

T, V are perspective

==> V, mT are perspective.

Now,

mV is inscribed in V, and V, mV are perspective.

V is inscribed in mT

V, mT are perspective [see above]

==> mT, mV are perspective

OED

Good Night from the rainy Athens

Antreas
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