Problem
GEO-B2-M10-P023 Symmedian Through Areas
In triangle \(ABC\), the median \(AM\) and cevian \(AD\) to side \(BC\) are isogonal, that is, \(\angle BAD=\angle MAC\) and \(\angle CAD=\angle MAB\). Prove that \(BD:DC=AB^2:AC^2\).
First express \(BD:DC\) through the areas of \(ABD\) and \(ACD\), then use equality of the areas of \(ABM\) and \(ACM\).
Since triangles \(ABD\) and \(ACD\) have the same altitude from \(A\), \(\frac{BD}{DC}=\frac{[ABD]}{[ACD]}\). Also, \(\frac{[ABD]}{[ACD]}=\frac{AB\cdot AD\sin\angle BAD}{AC\cdot AD\sin\angle CAD}=\frac{AB\sin\angle BAD}{AC\sin\angle CAD}\). The median \(AM\) divides the triangle into equal areas, so \(AB\sin\angle MAB=AC\sin\angle MAC\). By isogonality, \(\sin\angle BAD=\sin\angle MAC\), and \(\sin\angle CAD=\sin\angle MAB\). Therefore \(\frac{BD}{DC}=\frac{AB}{AC}\cdot\frac{\sin\angle MAC}{\sin\angle MAB}=\frac{AB}{AC}\cdot\frac{AB}{AC}=\frac{AB^2}{AC^2}\).
A strong mixed problem: areas, isogonality, and a side ratio.