Triangle Inequality Proof: $AD+DC \le AB+BC$ for Point D in Triangle ABC

[Jameson]

Thank you to Ackbach for this problem and to those of you who participated in last week's POTW!

Given a triangle $ABC$ and a point $D$ inside $ABC$, prove that $\overline{AD}+\overline{DC}\le \overline{AB}+\overline{BC}$.

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) caffeinemachine

Solution:

[sp]Extend line $AD$ such that it intersects with line $BC$ at point $E$. Use the triangle inequality twice:

$$\overline{AD}+\overline{DC}\le \overline{AD}+\overline{DE}+\overline{EC}$$

$$=\overline{AE}+\overline{EC}\le \overline{AB}+\overline{BE}+\overline{EC}$$

$$=\overline{AB}+\overline{BC}.$$

QED[/sp]