Given a triangle $ABC$, make it a point $D$ on the side $AB$. Show that $\overline {CD}$ is smaller than the length of one of the sides $BC$ and $AC$.
Ideas? The triangular inequality will not.
I wanted to try the theorem of the exterior angle and then apply a preposition that says "If two angles of a triangle are not congruent, then the sides who oppose these angles are different measures and the long side opposes mair angle".
We may assume without loss of generality that $CA\geq CB$. Then the triangle $ABC$ lies inside the circle $\Gamma$ with centre $C$ and radius $CA$. Assuming $CD>CA$, $D$ lies outside $\Gamma$, but that leads to a contradiction, since a circle is a convex set and the segment $AB$ lies inside $\Gamma$.
Another approach: assume that the projection of $C$ on the $AB$-line, say $H_C$, lies between $A$ and $B$. For any $P\in H_C B$, the length of $PC$ is between $CH_C$ and $CB$ by the Pythagorean theorem, and for any $P\in H_C A$, the length of $PC$ is between $CH_C$ and $CA$ for the same reason, hence for any $P\in AB$, $PC$ cannot exceed $\max(CA,CB)$. On the other hand, if the projection of $C$ on the $AB$-line lies outside the $AB$-segment, the map $P\to d(P,C)$ is a monotonic map on the $AB$-segment, and $PC\leq\max(CA,CB)$ still holds.
The triangle inequality--in Euclidean vector space--does work. Think of $A$, $B$, $C$, $D$ as vectors in $\mathbb R^d$, so that the Euclidean norm $||A-C||$ is the length of the side $AC$, and $||D-C||$ is the length of segment $DC$, and so on. By definition of $D$, we have $D=\lambda A + (1-\lambda) B$ for some $\lambda$ between $0$ and $1$. Then $$ \begin{align} ||D-C||&= ||\lambda (A-C)+(1-\lambda)(B-C)||\\ &\le \lambda||A-C|| +(1-\lambda)||B-C||\\&\le\max\{||A-C||,||B-C||\} \end{align}$$
Angles $ADC$ and $BDC$ are supplementary. Notice that at least one of then must be obtuse or both $90$.
If angle $BDC$ is obtuse,then $BC > CD$.(or if angle $ADC$ is obtuse)
If angle $BDC$ is $90$,then $BC$ is the hypotenuse.
