Originally Posted by
masters Prove: $\displaystyle (KF)^2=KN \cdot KC$
This follows easily from $\displaystyle \triangle NKF \sim \triangle FKC$
and setting up your corresponding proportional sides:
$\displaystyle \frac{NK}{FK}=\frac{KF}{KC}$
$\displaystyle (KF)^2=NK \cdot KC$
Prove: $\displaystyle KF=KE$
$\displaystyle (KE)^2=KN \cdot KC$ because a $\displaystyle \overline {KE}$ is a tangent and $\displaystyle \overline {KC}$ is a secant intersecting the circle at N. Justification: Secant-Tangent Theorem.
Using the previous conclusion and this, we use substitution to obtain:
$\displaystyle (KF)^2=(KE)^2$
$\displaystyle KF=KE$