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