Problem
GEO-B1-M04-P011 Isosceles Trapezoid and a Circle
#11
★★☆☆☆ Level 2 of 5
In trapezoid \(ABCD\), bases \(AD\parallel BC\), and the legs are equal: \(AB=CD\). Prove that points \(A,B,C,D\) lie on one circle.
In an isosceles trapezoid, base angles are equal. Then use the sum of opposite angles.
Since \(AB=CD\), the trapezoid is isosceles, so \(\angle DAB=\angle CDA\). From \(AD\parallel BC\), we get \(\angle DAB+\angle ABC=180^\circ\). Replacing \(\angle DAB\) by equal angle \(\angle CDA\), we obtain \(\angle ABC+\angle CDA=180^\circ\). The sum of opposite angles is \(180^\circ\), hence quadrilateral \(ABCD\) is cyclic.
A good first problem on proving cyclicity without heavy circle theory.