Problem
GEO-B2-M02-P017 A Tangent to a Circumcircle and a Ratio
#17
★★★★☆ Level 4 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 \(\frac{TB}{TC}=\frac{AB^2}{AC^2}\).
Use both the power of point \(T\) and the similarity \(\triangle TAB\sim\triangle TCA\).
By the tangent-secant theorem, \(TA^2=TB\cdot TC\). By the tangent-chord theorem, \(\angle TAB=\angle ACB\) and \(\angle TAC=\angle ABC\), so \(\triangle TAB\sim\triangle TCA\). From similarity, \(\frac{TB}{TA}=\frac{AB}{AC}\) and \(\frac{TA}{TC}=\frac{AB}{AC}\). Multiplying, we get \(\frac{TB}{TC}=\frac{AB^2}{AC^2}\).
A strong combination of Book 2 Module 1 and Module 2.