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\ x = 30^{0} \ exterior angle = 30 + 30 = 60^{0} Also D ADC is isosceles (given ) \ Ð ACD = y^{0} \ y + y + 60 = 180^{0} ( angles of D ADC ) \ y = 60^{0} \ D ADC is equilateral \ AC = DC But DC + ED ( given ) \ AC = ED Hence C. 
Test 4 Answer Explanation To Test 4
