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Explanation The applet attempts to suggest the following statement:

 On each side of ΔABC, construct a pair of isotomic conjugates: K, L on AB, M, N on CA and P, Q on BC. Assume that in each case the linear barycentric coordinates of K, M, P (and hence those of L, N, Q) are the same. (Which only means that, say, AK/AB = BP/BC = CM/CA.) Using KL, MN, PQ as bases form similar and similarly oriented isosceles triangles KLC', MNB', and PQA'. Then lines AA', BB', CC' are concurrent.

The forgoing statement generalizes Kiepert's theorem which follows when KL span the whole AB, etc.

In [Garfunkel, Stahl], the concurrency is observed and proven when the named segments coincide with the middle thirds of the corresponding sides and the constructed triangles are equilateral. We prefer to treat a more general theorem.

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### Proof

A proof follows directly from a theorem by D. Gale.

To use the theorem, we have to show that the segments AB' and AC' are isogonal conjugate as are the pairs CB', CA' and BA', BC'.

Since K and L are isotomic, AK/AB = LB/AB, so that LB/AB = BP/BC. Also, KL/AB = PQ/BC. Since triangles KLC' and PQA' are similar, we get another proportion: C'L/AB = A'Q/BC. Angles BLC' and BPA' are equal as being supplementary to equal angles. From which we conclude that triangles BLC' and BPA' are similar, implying that angles LBC' and PBA' are equal. Thus BA' and BC' are indeed isogonal conjugates.

For the pairs CB', CA' and BA', BC' the derivation is similar and we are in a position to apply D. Gale's theorem.

### References

1. J. Garfunkel, S. Stahl, The Triangle Reinvestigated, Am Math Monthly, Vol. 72, No. 1. (Jan., 1965), pp. 12-20  