# tangent circles radical axis proof

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!

• I'm a little confused. Isn't AN = BN and AM = BM? – fleablood May 18 '17 at 21:28
• @fleablood Possibly. Why would they be equal? – Parley May 18 '17 at 21:29
• @fleablood I thought so at first glance, but I think that requires $TB=AT$. All that's given is $T$ is between $A$ and $B$. – MichaelChirico May 18 '17 at 21:31
• If TA does not equal TA than the tangent lines will not meet at the same points. – fleablood May 18 '17 at 21:35
• *** Picture was wrong!! I fixed it! – Parley May 18 '17 at 22:23

## 2 Answers

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.

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 $AN + BM = AM + BN$.

• I appricate this but this is not what the proof is asking. How can this be used to show AM+BN=AN+BM – Parley May 18 '17 at 21:55
• Are you kidding me? AM = AN and BN = BM so AM + BN = AN + BM. – fleablood May 18 '17 at 21:57
• Hmm... from the OP’s diagram, it’s pretty clear that $AM\ne AN$. Did something change in the question since you posted this answer? – amd May 19 '17 at 0:04
• @amd yes! The picture was wrong originally! The picture that is up now is correct! – Parley May 19 '17 at 0:07