Circles
Circle
Allen Star Batch
Grade 11
Question:
Let $BD$ be the internal angle bisector of angle $B$ in triangle $ABC$ with $D$ on side $AC$. The circumcircle of triangle $BDC$ meets $AB$ at $E$, while the circumcircle of triangle $ABD$ meets $BC$ at $F$. If $AE = 3$, then $CF$ is equal to ______.
Step-by-Step Solution
Key Concept: Power of a point theorem applied to two intersecting circles yields relationships between chord segments.
Let $S_1$ be the circumcircle of triangle $BDC$ and $S_2$ be the circumcircle of triangle $ABC$. From power of point $A$ with respect to $S_1$: $AE \times AB = AD \times AC$, giving $AE = \frac{AD \times AC}{AB}$. From power of point $C$ with respect to $S_2$: $CF \times CB = CD \times CA$, giving $CF = \frac{CD \times CA}{AB}$. Therefore $\frac{AE}{CF} = \frac{AD \times CB}{AB \times CD}$. Given the ratio condition $\frac{AD}{CD} = \frac{AB}{BC}$, we find $AE = CF$, so $CF = 3$.
Correct Answer: 3