# Find the value of x.

Please see the diagram for the problem below ,

Calculate some known angles:

- ACB = 180-(10+70)-(60+20) = 20°
- AEB = 180-70-(60+20) = 30° Draw a line from point D parallel to AB, labeling the intersection with BC as a new point F and conclude:
- DCF ACB
- CFD = CBA = 60+20 = 80°
- DFB = 180-80 = 100°
- CDF = CAB = 70+10 = 80°
- ADF = 180-80 = 100°
- BDF = 180-100-20 = 60°

- ADF BFD
- AFD = BDF = 60°
- DGF = 180-60-60 = 60° = AGB
- GAB = 180-60-60 = 60°
- DFG (with all angles 60°) is equilateral
- AGB (with all angles 60°) is equilateral . CFA with two 20° angles is isosceles, so FC = FA

- ACG CAE
- FC-CE = FA-AG = FE = FG
- FG = FD, so FE = FD . With two equal sides, DFE is isosceles and conclude:
- DEF = 30+x = (180-80)/2 = 50

Thanks & Regards ,

Harshit Tyagi

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