Proving a Triangle is Equilateral if and only if it is Equiangular

Question

Link to the modified version of Euclid's Elements referenced and used in the making of this question

Just as a quick disclaimer, this is the first post I have made here, so chances are I will need to improve aspects (or all) of it. Additionally, I am very new to mathematical proofs in general, which is probably responsible for my confusion that led to this question. All the propositions I mention in this question are from Book 1 of Euclid’s Elements.

Hello, I am currently reading through a modified version of Euclid’s Elements (I have attached a link to the full document above), which has a variety of exercises in related to specific propositions. One exercise is to prove a corollary of Euclid’s 5th proposition in Book 1. The exercise asks you to prove the following statement:

A triangle is equilateral if and only if it is equiangular.

So far, I have found out that, since the corollary includes “if and only if” in its wording, I would have to prove that an equilateral triangle must be equiangular, in addition to also proving either the inverse or converse of this statement. As of right now, I have largely gotten stuck when proving anything related to this corollary. If possible, I would like to prove this statement exclusively using Euclid’s definitions and propositions 1-5 (including 5). My current idea is the following (See image 1 for a construction of this idea):

1. Draw an equilateral triangle (using the 1st proposition).
2. Draw the triangle’s 3 medians to get its centroid.
3. This is about where I have gotten stuck, my idea, if I could prove that $AG = BG,$ is that, by using proposition 5, $\angle GAD = \angle GBD.$
4. From there I would then apply this process to the other triangles, though I am unsure as to how I would prove that those pairs of angles in isosceles triangles equal all of the other pairs respectively (i.e., how I could prove that $\angle GAD$ and $\angle GBD$ equal $\angle CAG$ and $\angle CBG$?).

When it comes to proving the other statements required (either the converse or the inverse) I have come even less close to the solution. In one of my attempts to prove the converse statement (If a triangle is equiangular, it is equilateral), I attempted something similar to my other idea above, where I follow the following sequence of steps (see image 2 for a construction of this idea):

1. Draw an equilateral triangle (using the 1st proposition).
2. Draw the triangle’s 3 angle bisectors to get its incenter.
3. From there, I had the idea to draw a circle at point H of radius HA (the line segment HA wasn’t drawn in the image).
4. My idea was to prove that $HA = HB,$ via proving that both are radii of circle $H.$ To do so, I attempted a proof by contradiction, wherein I assumed $HA < HB$ (which would mean $HB$ isn’t a radii of circle $H$), then attempted to obtain a contradiction.
5. I first cut off $HI$ from $HB,$ such that $HA = HI$ (proposition 3), then attempted to obtain a contradiction to prove that $HB$ is actually equal to $HA,$ but once again got stuck.

I would heavily appreciate help of any kind, but mainly I would like to find out where I am making a mistake in finding a proof for the corollary. I believe that I may have been over-complicating things with my attempted proofs, but if there is a way to add on to my work to complete the proof, I would appreciate any methods, tips, or hints you can give me in doing so.

Image 1

Image 2