Problem
GEO-B3-M05-P003 Symmedian Criterion
In triangle \(ABC\), cevian \(AS\) meets \(BC\) at \(S\). Prove that \(AS\) is the \(A\)-symmedian if and only if \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).
C. Hint 1. Compare \(AS\) with its isogonal median.
D. Hint 2. For any cevian use \(\frac{BX}{CX}=\frac{AB\sin\angle BAX}{AC\sin\angle XAC}\).
For an arbitrary cevian \(AX\), the sine rule in triangles \(ABX\) and \(ACX\) gives \(\frac{BX}{CX}=\frac{AB\sin\angle BAX}{AC\sin\angle XAC}\).
If \(AS\) is the isogonal image of the median \(AM\), then \(\angle BAS=\angle MAC\), \(\angle SAC=\angle BAM\). Since \(BM=CM\), the median gives \(\frac{\sin\angle BAM}{\sin\angle MAC}=\frac{AC}{AB}\). Substitution yields \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).
Conversely, if this ratio holds, the same cevian formula shows that the isogonal line to \(AS\) divides \(BC\) equally, so it is the median. Hence \(AS\) is the symmedian.
This criterion will be used as the working definition.