Is this solution credible

In triangle ACB:

Triangle ACD: I dropped a perpendicular (DE) down to line AC, creating for Triangle ADE and found line DE by taking the hypotenuse AD=100, thus the 30* angle ADE would have the line AE=50. the (100^2)-(50^2)= (approximately) 83^2 for line/side DE. W/angle DEC being 90*, I presumed angle DCE would be 30*, thus making line DC 166 , twice the size of the side opposite the 30* angle.
Then I went to Triangle ABD, and presumed to make a 90* angle at ADF with AF being the hypotenuse of Triangle ADF. Since ADF =90* and DAF=60*, then angler AFD=30*. By rule, since the opposite side of the 30* angle=100, then the hypotenuse FA=200. Line DF=173
From point F I dropped a perpendicular down to line BDC. Now, just concerned with triangle DFG. IF angle DFA=30*. angle DFG=90*. Angle FGD=60* . Angle GDF=30*. Using the equation 173/3^0.5=99.9 (rounded up to 100).
Triangle FBG:
Since angle AFD=30*, angle DFG=90*, then angle GFB=30* and angle FBG=60* and angle BGF=90*
Since line FG=100 and is opposite the 60* angle, then by dividning 100/3^5=57
Triangle DFG
Angle FGD=90*. Line DF=173. Line FG=100. By rule, 173^2-100^2=141^2

See image https://pbs.twimg.com/media/EMUBTxxX0AEiKBx?format=png&name=small

57+141+166=364