In trapezoid ABCD with bases AD and BC, the diagonal AC is perpendicular to the lateral side of CD

| September 10, 2021 | education

In trapezoid ABCD with bases AD and BC, the diagonal AC is perpendicular to the lateral side of CD. It is known that AB = BC = CD. Find the angle BAC.

Let’s draw a diagonal BD. Triangles ABD and ACD are equal, since AB = CD, side AD is common to them, and angle BD = CDA, as angles at the base of an isosceles trapezoid. Triangles are equal on two sides and the angle between them. Triangle ACD is rectangular, therefore ABD is rectangular with right angle ABD.

Since AB = BC, the triangle ABC is isosceles, the angle BAC = BCA.

In an isosceles trapezoid, the diagonals are divided at the point of intersection into equal segments, then the BOS triangle is isosceles, the angle BCO = CBO. BCA angle = CAD as criss-cross angles.

Let the angle BAC = X0, then the angle BCA = СBО = CAD = X0. In the trapezoid, the sum of the angles at the lateral side is 180, then the angle BAD + ABC = CAD + BAC + ABD + CBO = X + X + 90 + X = 180.

3 * X = 180 – 90 = 90.

X = BAC = 90/3 = 30.

Answer: The BAC angle is 30.



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