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Prove If |A-B|< (d/2) and |B-C|< (d/2) then |A-C| < d
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A,B,C are 3 points right?
i mean i guess they could be. Doesnt really specify. I'm just learning proofs in class right now so it doesn't technically state anything
i think A,B,C are 3 points, and |A - B| are distance between A and B and likewise for others 2 cases you can consider: first is all 3 points lie on a straight line...so distance A to B to C, you can draw a line and see that's obvious. second if they are not on straight line, they form a triangle, in a triangle the length of 2 side is always greater than the length of one side so: |A - C| < |A - B| + |B - C| < d/2 + d/2 = d, therefore, |A - C| < d
i mean sum of length of 2 sides, ...forgot the sum in there
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