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Prove that AX + XB = AB for any X on AB. 

(Hint: examine all cases)

Hope u can help!


The TRIANGLE INEQUALITY states that for any three points A, B, X in the Euclidean plane, AX + XB >= AB, with equality if and only if X lies on the segment AB (somewhere between A and B or at the endpoints).

The claim you wish to prove is not true unless you introduce "directed distances," where PQ = -QP for any two points P and Q. Note that directed distances make sense only among points on the same line. Then the method of proof depends on one's axiom system. The triangle inequality (of which your problem is a special case) is often taken as an AXIOM. The easiest treatment would be to introduce coordinates: let A = (0, 0) be the origin, B = (1, 0) be the unit point on the x-axis, and let X = (x, 0) be an arbitrary point of the x-axis. Using algebra you can deal with all possibilities in one step since x can be any number, positive or negative or zero: AX = x, XB = 1-x, and AB = 1.

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