An equilateral triangle is partitioned into smaller equilateral triangular pieces. Prove that two of the pieces are the same size.

Question

Here's the Problem (Its an NZMO $2019$ question):

An equilateral triangle is partitioned into smaller equilateral triangular pieces. Prove that two of the pieces are the same size.

As well as being a pretty weird Olympiad question to grasp the meaning of, I also do not know how to finish this question.

Here is my beginning:

Based on the information on the question, we can logically state that there are only $3$ types of vertex for the equilateral triangle.

$•$ Type $A:$ incident with only $2$ edges at $60°$.

$•$ Type $B :$ incident with exactly $4$ edges at angles $60°, 60, 60°, 180°$

$•$ Type $C :$ incident with exactly $6$ edges forming $6$, $60°$ angles.

There can be no other types of vertex, because all the angles must be $60°$ or $180°$ or $300°$ in the corners of the original large triangle.

I have only so far been able to solve this question up to here. If anybody could help me to finish this question, I would appreciate it. Thank you very much.

Edit: I would like the proof to be different from the one Beni Bogosel suggested for another similar question (it is on Math Stack Exchange too): Dissecting an equilateral triangle into equilateral triangles of pairwise different sizes

Beni suggests that proceeding in a recursive way:

"The triangle $A_1B_1C_1$ and find one smallest triangle pointing downwards with side contained in $B_1C_1$ $...$ this triangle has two neighbor triangles and one of them is smaller, which we denote by $A_2B_2C_2$, and so on $...$ continue this procedure indefinitely because all the triangle sides are different."

This is the last question (Q$5$ of an invitational paper) and they began it in my way above. I would like some to continue with this solution (obviously not their approach as I don't really understand it). Thanks.