The diagonal AC of trapezoid ABCD is equal to the lateral side of CD, is perpendicular to it and divides the angle BAD in half. Find the corners of the trapezoid.
Consider a triangle ACD.
Since by the condition of the problem AC = CD and AC, CD are perpendicular,
then the triangle ACD is right-angled and isosceles.
Consequently, the angles at its hypotenuse are equal to 45 degrees.
Hence, CDA = 45 ° and BCD = 180 ° – 45 ° = 135 °.
Since AC is the bisector of the BAD angle, CAB = CAD = 45 °.
Hence, BAC = 2 * CAD = 2 * 45 ° = 90 °.
Then ABC = 180 ° – BAC = 180 ° – 90 ° = 90 °.
So, the angles A, B, C, D of the trapezoid are equal: 90 °, 90 °, 135 °, 45 °.
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