While trying to provide a proof for this question, I stumbled upon a theorem that I have probably seen before:

Theorem. If line segments joining corresponding vertices of two similar triangles in the same orientation (not reflected) are split into equal proportions, the resulting points form a triangle similar to the original triangles.

enter image description here

In the diagram above, $RZ: ZU = TA_1:A_1W = SB_1:B_1V$, and the purple triangle is similar to the orange triangles. Note that the lines does not need to be parallel or concurrent, so the triangles are not necessarily in perspective or share any parallel sides. This theorem can also be extended easily to polygons.

I am confident that this is indeed a theorem, but I am having a hard time recalling its name, and searching for geometric theorems is practically impossible.

For those who are interested, a solution to the linked question is immediate using this theorem and a simple reflection:

enter image description here

since $CH:HE = B'G:GB = AI:ID = 2:1$ where $B'$ is the reflection of $B$ across the line $AC$.


Let us use complex numbers geometry.

Two triangles $T=(z_1,z_2,z_3)$ and $T'=(z'_1,z'_2,z'_3)$ are similar iff there exists a similitude $Z=az+b$ (with $a,b \in \mathbb{C}$ ) sending the first one onto the second one.

Therefore if a third triangle $T''$ is for example at the two-thirds from $T$ and $T'$ with the relationship described above, we can write :

$$\begin{cases}z"_1&=&\frac13 z_1+\frac23 z'_1&=&\frac13 z_1+\frac23 (az_1+b)&=& \frac{2a+1}{3}z_1+\frac23 b\\ z"_2&=&\frac13 z_2+\frac23 z'_2&=&\frac13 z_2+\frac23 (az_2+b)&=& \frac{2a+1}{3}z_2+\frac23 b\\ z"_3&=&\frac13 z_3+\frac23 z'_3&=&\frac13 z_3+\frac23 (az_3+b)&=& \frac{2a+1}{3}z_3+\frac23 b\end{cases}$$

exhibiting a similitude

$$Z=\frac{2a+1}{3}z+\frac23 b$$

sending $T$ onto $T''$, therefore establishing that $T$ and $T''$ are similar.

Remark: Of course, what we have done with coefficients $\frac13$ and $\frac23$ can be done with any other "weighting" on triangles $T$ and $T'$.

  • $\begingroup$ Any comment ... ? $\endgroup$ – Jean Marie Jan 21 at 9:17

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