Problem
GEO-B1-M04-P024 Isosceles Trapezoid and Diagonals
In isosceles trapezoid \(ABCD\), bases \(AD\parallel BC\), \(AD=18\), \(BC=10\). Diagonals \(AC\) and \(BD\) meet at \(O\). Prove that \(AO=DO\), and find \(AO:OC\).
First prove that the diagonals of an isosceles trapezoid are equal. Then use the diagonal ratio in a trapezoid.
In an isosceles trapezoid, base angles are equal, so triangles \(DAB\) and \(ADC\) are congruent by two sides and the included angle: \(AD\) is common, \(AB=CD\), and the base angles at \(AD\) are equal. Therefore \(BD=AC\). Since the diagonals of a trapezoid are divided in the ratio of the bases, \(AO:OC=DO:OB=AD:BC=18:10=9:5\). The diagonals \(AC\) and \(BD\) are equal and divided in the same ratio \(9:5\), so their larger parts are equal: \(AO=DO\). Answer: \(AO:OC=9:5\).
Final problem of the module: it combines isosceles trapezoids, equal diagonals, and similarity.