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

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SUMMARY

The discussion centers on proving the triangle inequality for a point D inside triangle ABC, specifically that $\overline{AD} + \overline{DC} \le \overline{AB} + \overline{BC}$. The proof involves extending line AD to intersect line BC at point E and applying the triangle inequality twice. The final conclusion confirms that the sum of the lengths from point D to the vertices A and C does not exceed the sum of the lengths of sides AB and BC. Participants, including member caffeinemachine, successfully contributed to the solution.

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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 $\overline{AD}+\overline{DC}\le \overline{AB}+\overline{BC}$.
 
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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]
 

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