Problem
GEO-B2-M01-P017 A Cyclic Trapezoid
#17
★★★★☆ Level 4 of 5
In trapezoid \(ABCD\), bases \(AD\parallel BC\), and points \(A,B,C,D\) lie on one circle. Prove that \(AB=CD\).
Compare the base angles at \(BC\), using parallelism and the sum of opposite angles in a cyclic quadrilateral.
Since \(AD\parallel BC\), angles \(\angle DAB\) and \(\angle ABC\) sum to \(180^\circ\). Since \(ABCD\) is cyclic, \(\angle DAB+\angle DCB=180^\circ\). Therefore \(\angle ABC=\angle DCB\). In a trapezoid, equality of the angles at one base implies equality of the legs: for example, drop the altitudes from \(A\) and \(D\) to \(BC\) and obtain two congruent right triangles. Hence \(AB=CD\).
Not a school property to memorize, but a proof through two angle reasons.