In an isosceles trapezoid ABCD, the angle between the diagonal AC and the base AB is 40 degrees and AD = DC. Find the corners of the trapezoid
Angle DCA = CAB = 40 as intersecting angles at the intersection of parallel straight lines DC and AB secant AC.
Since, according to the condition, AD = DC, then the triangle ADC is isosceles, which means the angle DAC = DCA = 40.
Then the angle DАВ = DАС + BAC = 40 + 40 = 80.
Since the sum of the angles at the lateral sides of the trapezoid is 180, the angle ADC = 180 – 80 = 100.
Angle DCB = ADC = 100, angle ABC = DAB = 80, since the angles at the bases are equal.
Answer: The angles of the trapezoid are 80, 100, 80, 100.
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