Two Circles with centers $O_1$ and $O_2$ are tangent exterior at point T. Let points A and B be on the common tangent line through T such that T is between A and B. The tangent lines from A and B to circle $O_1$ meet at point M and the tangent line from A and B to circle $O_2$ meet at point N.
Prove that AM+BN=AN+BM
I know that ATB is the radical axis of the two circles. I am assuming the proof is about power of the point of points M, N, A, and B. Any help would be appreciated!
The diagram in fleablood’s answer is the key to this proof. Let $P$ and $Q$, respectively, be the intersections of $\overline{AM}$ and $\overline{BM}$ with circle $O_1$, and $R$, $S$ be the intersections of $\overline{AN}$ and $\overline{BN}$, respectively, with circle $O_2$. We have $AP=AT=AR$, $MP=MQ$, $BQ=BT=BS$ and $NR=NS$. We also have $AM=AP+NP$, $BM=BQ+MQ$, $AN=AR+NR$ and $BN=BS+NS$. The rest should be obvious.
http://staff.argyll.epsb.ca/jreed/math20p/circles/images/tangentProof.png
So it's very clear that $MA = MB$ and $NA= NB$.
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Okay. $\angle TMA \cong \angle TMB$ because $MA$ and $MB$ are both tangents to the same circle. So $\triangle TMA \cong \triangle TMB$ by Angle ($\angle TMA = \angle TMB$) Side ($AT=AT$) Angle ($\angle MTA = \angle MTB$). So $AM = AN$.
Similarly $BN = BM$.
So $AN + BM = AM + BN$.
