Problem
GEO-B2-M02-P021 A Tangent and the Symmedian Ratio
#21
★★★★★ Level 5 of 5
The tangent to the circumcircle of triangle \(ABC\) at \(A\) meets line \(BC\) at point \(T\), with \(T\) outside segment \(BC\). Prove that \(TB:TC=AB^2:AC^2\), and then find \(TB:TC\) if \(AB=9\), \(AC=6\).
First prove the general ratio, then substitute the lengths.
By the result proved above for a tangent to the circumcircle, \(\frac{TB}{TC}=\frac{AB^2}{AC^2}\). For completeness: it follows from the similarity \(\triangle TAB\sim\triangle TCA\), obtained by the tangent-chord theorem. For \(AB=9\), \(AC=6\), we get \(TB:TC=9^2:6^2=81:36=9:4\).
This is a strong olympiad lemma; the word “symmedian” may be mentioned orally, but the theory need not be overloaded.