In an isosceles trapezoid, the acute angle is 60 °. Prove that the smaller base is equal

| February 24, 2021 | education

In an isosceles trapezoid, the acute angle is 60 °. Prove that the smaller base is equal to the difference between the larger base and the side.

Let’s construct the heights of the ВН and СK of the trapezoid ABCD.

The ВНKС quadrangle is a rectangle, then НK = BC, and the triangles ABN and CDK are rectangular.

In right-angled triangles ABН and CDK, the hypotenuse AB and CD are equal, since the trapezoid is isosceles, and, accordingly, the angle BAН = CDK, then the triangles are equal in hypotenuse and acute angle, which means AH = DK = (AD – НK) / 2 = (AD – BC) / 2.

In a right-angled triangle ABН, the angle ABН = (90 – 60) = 30, then the leg AH lies opposite the angle 30, and therefore AB = 2 * AH = 2 * (AD – BC) / 2.

AB = (AD – BC).

BC = AD – AB, as required.



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