Question

find the value of x-​

Answer 1

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• 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 and ∆ACB

• CFD = CBA = 60+20 = 80°
• DFB = 180-80 = 100°
• CDF = CAB = 70+10 = 80°
• ADF = 180-80 = 100°
• BDF = 180-100-20 = 60°

Draw a line FA labeling the intersection with DB as a new point G and conclude:

∆ADF and ∆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

Draw a line CG, which bisects ACB and conclude:

• 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

• x = 20°

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Answer 2

Answer:

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 and ∆ACB

CFD = CBA = 60+20 = 80°

DFB = 180-80 = 100°

CDF = CAB = 70+10 = 80°

ADF = 180-80 = 100°

BDF = 180-100-20 = 60°

Draw a line FA labeling the intersection with DB as a new point G and conclude:

∆ADF and ∆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

Draw a line CG, which bisects ACB and conclude:

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

x = 20°