# Sum of distances from triangle vertices to interior point is less than perimeter?

Let $M$ be a point in the interior of triangle $ABC$ in the plane. Prove $AM+BM+CM<AB+BC+CA$.

The above question was posed to someone I know who is taking high-school Euclidean geometry. I'm not sure what theorems she can rely on in her proof (though they all follow from Euclid's axioms anyway), but I do know that she does not use trigonometry at all. She turned to me (a mathematician by training) for help; and I can't seem to prove it. So I turn to you all for a proof (using facts from high-school geometry only).

One thing I can prove is that $\sup(AM,BM,CM)<\sup(AB,BC,CA)$. Indeed, say the longest of the interior segments is $\overline{AM}$. Drop an altitude (perpendicular) from $A$ to the point $D\in\overline{BC}$, and consider the side — $\overline{AB}$ or $\overline{AC}$ — such that $\overline{AM}$ lies between $\overline{AD}$ and that side. (If $\overline{AM}\subset\overline{AD}$, consider either $\overline{AB}$ or $\overline{AC}$.) Say it's $\overline{AB}$. Then examining right triangle $ADB$ shows easily that $AM<AB$. However, I can't seem to prove that each of the three sides can be used in turn for one of the interior segments in that proof — which would suffice for the problem above.

Another idea I had was to prove that angle $AMB$ is strictly larger than angle $ACB$ (and likewise for the other two angles) and to use that to prove the claim. But I can't seem to do either: neither to prove the inequality of angle measures, nor, assuming that inequality, to prove the claim sought.

Any help would be much appreciated — again, using high-school geometry only. I suspect there's a simple proof I'm not seeing.

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The claim is relatively straightforward to prove for $M$ on the perimeter. Then the fact that the sum is convex as $M$ moves on any line from the perimeter to the opposite vertex shows that the claim also holds in the interior. Thus it would suffice if you could formulate and prove this convexity in geometrical terms. – joriki Jun 7 '13 at 6:38

The method is to prove $AC+BC>AM+BM$:

Extend $AM$, let $ME=MB \implies \angle MBE=\angle MEB$, $AM$ cross $BC$ at $F$ (because $M$ is inside of $\triangle ABC$).

If $E$ is on $MF$, then $AF\ge AE=AM+ME$. $BC>CF \implies BC+AC> FC+AC>AF \ge AE=AM+BM$,

If $E$ is on the extension $MF$, $\angle MBE> \angle CBE, \angle BEC>\angle MEB =\angle MBE > \angle CBE$, so in $\triangle CEB$, $BC>CE, \implies BC+AC>AC+CE>AE=AM+BM$.

So we've proven $AC+BC>AM+BM$. For the same reason, $BC+AB>AM+MC$, $AB+AC>BM+MC$.

Finally, we have $AC+BC+AB>AM+BM+CM$.

edit1: more simple way:

$AC +CF> AF =AM +MF, MF + BF > BM \\ \implies AC+CF+BF+MF > AM+BM+MF \\ \implies AC+BC> AM+BM$

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+1; very nice; many thanks. – msh210 Jun 7 '13 at 7:57
my pleasure if it helps! – chenbai Jun 7 '13 at 12:13