thanks keebler303 for getting back so quick. already many valuable lessons learned.
I made a note that it is better to start from 0 and work from there to the outer edge of the cut and from there to a path tool pivot point. Like you showed in case of D for Y. 0 + 0.6900/2 gets us to the outermost upper edge of the bigger cut which is on the right, - 0.0950 drops us to the upper edge of the cut on the left, -0.5000 positions us at the lower edge of the cut and +0.1875 moves our point to a desired location which is D.

what was a bit counter-intuitive to me at first was the 0.2180 - 0.1875 = 0.0305. I added a hand scribble maybe it might help others to better grasp it as well.. i hope i got it right.

Attachment 230172

heaving squared that it gets easier. so we have

Ye is the same as Yd so we have Ye = - 0.0625, Xe on the other hand form the outer left most edge of the smaller cut is 0-3.5120 + 0.1875(tool radius) + 0.0305 (calculated delta between big radius and tool radius, same deal like between C and D) = -3.2940 so overall E ( -3.2940,-0.0625)
Yf is the same as Yc = -0.0320, Xf going from outer left most edge of the cut is -3.5120 + 0.1875 (tool radius) = -3.3245, so F (-3.3245, -0.0320)

i'm going to post it now, so not to loose it, and continue further on. did i got it right? if any is off, could you just point it out which i will try to figure it out..