Problem
GEO-B1-M06-P024 Equal Areas at the Diagonals of a Trapezoid
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In trapezoid \(ABCD\), bases \(AD\parallel BC\), and the diagonals meet at point \(O\). Prove that \(S_{AOB}=S_{COD}\).
Compare areas \(ABD\) and \(ACD\), then subtract the common part.
Triangles \(ABD\) and \(ACD\) have common base \(AD\). The heights from \(B\) and \(C\) to \(AD\) are equal because \(BC\parallel AD\). Therefore \(S_{ABD}=S_{ACD}\). Subtract the common triangle \(AOD\) from both areas: we get \(S_{AOB}=S_{COD}\).
Classic area chasing: equal large areas minus a common part.