Problem
GEO-B2-M04-P006 Diagonals of a Trapezoid
#6
★★★☆☆ Level 3 of 5
In trapezoid \(ABCD\), bases \(AD\) and \(BC\) are parallel. Diagonals \(AC\) and \(BD\) meet at \(O\). Prove that \(\frac{AO}{OC}=\frac{DO}{OB}\).
Consider triangles \(AOD\) and \(COB\).
Angles \(\angle AOD\) and \(\angle COB\) are vertical. Since \(AD \parallel BC\), we get \(\angle ADO=\angle CBO\). Hence \(\triangle AOD \sim \triangle COB\). From similarity, \(\frac{AO}{OC}=\frac{DO}{OB}=\frac{AD}{BC}\), in particular \(\frac{AO}{OC}=\frac{DO}{OB}\).
Emphasise that this is a homothety with centre \(O\) sending one base to the other.