Problem
GEO-B1-M04-P015 Diagonals Divided Like the Bases
#15
★★★☆☆ Level 3 of 5
In trapezoid \(ABCD\), bases \(AD\parallel BC\), \(AD=21\), \(BC=14\). The diagonals meet at \(O\). Find \(AO:OC\) and \(DO:OB\), with justification.
Prove that triangles \(AOD\) and \(COB\) are similar.
Triangles \(AOD\) and \(COB\) are similar: \(\angle AOD=\angle COB\) as vertical angles, and \(\angle ADO=\angle CBO\) because \(AD\parallel BC\). Therefore \(\frac{AO}{OC}=\frac{DO}{OB}=\frac{AD}{BC}=\frac{21}{14}=\frac{3}{2}\). Hence \(AO:OC=DO:OB=3:2\).
Connects the similarity module with trapezoids.