Kiepert's Triangles Graduate to Ears of Arbitrary Shape

What is this about?


On the side of $\Delta ABC$ construct similarly oriented similar triangles $ABN_{c},$ $BCN_a,$ and $CAN_b.$

Similar ears on sides of a triangle

Prove that the centroids of triangles $ABC$ and $N_{a}N_{b}N_{c}$ coincide.


I think in complex numbers the proof is exceptionally simple. In affine geometry, though, it is no more difficult.


Let in a complex plain, the vertices of $\Delta ABC$ correspond to complex numbers $\alpha,$ $\beta,$ $\gamma.$

Similar ears on sides of a triangle

Then, if $i^{2}=-1$,

$N_{a}=\gamma + t(\beta -\gamma),$
$N_{b}=\alpha + t(\gamma -\alpha),$
$N_{c}=\beta + t(\alpha -\beta),$

for some complex $t$ that determines a base angle and the corresponding side length of the formed triangles.

The centroid of $\Delta ABC$ is given by $(\alpha +\beta +\gamma)/3$ while that of $\Delta N_{a}N_{b}N_{c}$ is

$\displaystyle \begin{align} \frac{[\gamma + t(\beta -\gamma)] + [\alpha + t(\gamma -\alpha)] + [\beta + t(\alpha -\beta)]}{3} \\ &= \frac{\alpha +\beta +\gamma}{3}. \end{align} $


I am sincerely grateful to Grégoire Nicollier who has kindly pointed out that the statement proved for Kiepert's isosceles triangles has an immediate extension to similar triangles of arbitrary shape, as above.

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