Suppose $ABC$ is any triangle and $BE$ is any line from the vertex $B$ to a point $E$ lying inside the segment $AC$. Let $D$ be any point on $BE$. I would like to verify the following: regardless of how we have drawn the picture above, it is true that $$ AB+BC\geq AD+DC. $$ I tried playing around with triangle inequalities but nothing came out of that. Haven't done geometry in many years so not surprised that I blow. Could you please help? Thanks.
On
Why not decompose direct connections in a vector notational way: $$ AB = AE + EB $$ Then it's simple to say $$ AB + BC = AE + EB + BE + EC \\ AD + DC = AE + ED + DE + EC $$ And now use $AB = BA$ for all $A, B$, so in the end $$ AB + BC \geq AD + BC \\ \Leftrightarrow EB \geq ED $$ which is certainly true by construction.
The locus of the points $P$ of the plane such that $AP+PC=AB+BC$ is the ellipse $\Gamma$ through $B$ with foci in $A$ and $C$. Since such ellipse is convex, for any point $Q$ inside $\Gamma$ we have $AQ+QC < AB+BC$.