In a rectangular trapezoid ABCD, the smaller base is perpendicular to the lateral side of AB. The AC diagonal is equal to the CD side, ∠CAD = 40. find all the corners of the trapezoid.
Since, by condition, the diagonal AC is equal to the lateral side of CD, then the triangle ACDD is isosceles, then its angles at the base of AD are equal.
Angle CAD = CDA = (180 – ACD) / 2 = (180 – 40) / 2 = 140/2 = 70.
The ACB angle is equal to the CAD angle as the intersecting angles at the intersection of parallel lines BC and AD secant AC. Then the angle ВСD = АСВ + АСD = 70 + 40 = 110.
The angles BAD and ABC are straight, since the trapezoid is rectangular.
Answer: The angles of the trapezoid are equal: 90, 90, 110, 70.
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