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

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In the discussion on proving the triangle inequality for point D inside triangle ABC, the key argument involves extending line AD to intersect line BC at point E. By applying the triangle inequality, it is shown that the sum of segments AD and DC is less than or equal to the sum of segments AB and BC. The proof is structured through a series of inequalities that connect the lengths of the segments. The solution provided by member caffeinemachine successfully demonstrates the required relationship. This proof reinforces the fundamental properties of triangles and the triangle inequality theorem.
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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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