Let the given rectilinear angle be that by BAG. It is, in fact, required to bisect it. (diagram 2) Let there by taken on AB a point as happens, D, (diagram 3) and let an equal, AE, to AD be taken away from AG, (diagram 4) and let DE be joined, (diagram 5) and let there be constructed on DA an equilateral triangle, DEZ, (diagram 6) and let AZ be joined. I say that the angle by BAG has been bisected by straight-line AZ. (diagram 7) For since AD is equal to AE, and AZ is common, in fact, two, DA, AZ, are respectively equal to two, EA, AZ. And a base, DZ is equal to a base, EZ. (diagram 8) Therefore, an angle, that by DAZ, is equal to an angle, that by EAZ. Therefore, the given rectilinear angle, that by BAG, has been bisected by straight line AZ, just what it was required to make.

Can u explain this to me(ignore the diagrams)
its euclidean geometry