Consider right \(\triangle ABC\) with \(\angle C=90^\circ\), \(AB=10\), and \(AC=8\). \(CP=CQ=2\). \(BQ\) and \(AP\) meet at \(R\). \(CR\) meet \(AB\) at \(S\). \(QP\) meet \(AB\) extension at \(T\). Find \(TS\).
Solution
Answer: \(24\)
The points \(A,S,B,T\) form a harmonic bundle, so \(\frac{AS}{BS}=\frac{AT}{BT}\). Alternatively, use Ceva’s and Menelaus’s theorems in \(\triangle ABC\).
Ceva’s theorem gives \(\frac{AS}{BS}=\frac32\). Since \(AB=10\),
\[\frac{10+BT}{BT}=\frac32,\qquad BT=20.\]
Also \(BS=4\), so \(TS=BT+BS=20+4=24\).