Problem
GEO-B2-M04-P014 A Trapezoid Criterion via Diagonals
#14
★★★★☆ Level 4 of 5
In quadrilateral \(ABCD\), diagonals \(AC\) and \(BD\) meet at \(O\). It is known that \(\frac{AO}{OC}=\frac{DO}{OB}\). Prove that \(AD \parallel BC\).
Prove that triangles \(AOD\) and \(COB\) are similar.
Triangles \(AOD\) and \(COB\) have vertical angles at \(O\). By the condition, \(\frac{AO}{OC}=\frac{DO}{OB}\), so the sides adjacent to these angles are proportional. Therefore \(\triangle AOD \sim \triangle COB\). Hence \(\angle DAO=\angle BCO\). These are alternate interior angles for lines \(AD\) and \(BC\), so \(AD \parallel BC\).
This is already an olympiad move: proving parallelism through a ratio of diagonal parts.