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Problem on its own (Posted on 2017-03-05) Difficulty: 3 of 5
Take a triangle, ABC, and an arbitrary point, D, in its interior.
How can we prove that AD + DB < AC + CB?

The fact seems obvious, but when the problem is presented on its own, outside of a textbook or some course of study, we have no hint as to what technique to use to prove it.
Construct an equation?
Apply the Pythagorean theorem?

I trust the flooblers to find a formal proof...

No Solution Yet Submitted by Ady TZIDON    
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Solution So simple! Comment 1 of 1
Extend AD to meet side BC at E.  ABC is then partitioned into triangles ACE, ABD, and BDE.

By the triangle inequality we have AC+CE>AE and DE+EB>DB. Then: AD+DB < AD+(DE+EB) = (AD+DE)+EB = AE+EB < (AC+CE)+EB = AC+(CE+EB) = AC+CB

Removing the intermediate steps leaves AD+DB < AC+CB, the desired result.

The surprising thing is how easy it is to overlook such an elementary approach!  My first thought was to draw CD and extend it to intersect AB, but that got trigonometry involved rather quickly.

  Posted by Brian Smith on 2017-05-05 11:49:11
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