circle with two tangent lines - how do I show that AN=MF?

Question

hello.

given that AD=DE=EF and that AB,BF are Tangent Line to the circle, how do I show that AN=MF ?

I thought about connecting DM , NE ,NM and to show that triangle BND is similar to ABF , using that BN=BM (because AB and BF are Tangent Line to the circle) , but then I am stuck. any help please?

Answer 1

Say $AD = DE = EF =x$.

By the power of the point $A$ we have $$AN^2 = AD\cdot AE = 2x^2$$ and the same with respect to $F$: $$FM^2 = FD\cdot FE = 2x^2$$

and thus a conclusion.

Answer 2

As shown above, let $O$ be the center of the circle and $G$ be the midpoint of $DE$. Join $O$ to $G$, $N$ and $M$. Then since $OE = OD = r$, where $r$ is the radius of the circle, $\triangle ODE$ is an isosceles triangle. Thus, the perpendicular bisector of $DE$ goes through $O$, so you have $\angle OGF = \angle OGA = 90^{o}$.

Also, $AG = GF$ and $OG$ is a common side, so by $SAS$ you have that $\triangle OGF$ and $\triangle OGA$ are congruent to each other. Thus, $OA = OF = x$. Also, since $N$ and $M$ are points of tangent lines to the circle, you also have $\angle ONA = \angle OMF = 90^{o}$. Using this plus $ON = OM = r$ gives by the Pythagorean theorem that $AN = MF = \sqrt{x^2 - r^2}$.