In any triangle difference of lengths of any two
two sides is less than the length of third side.
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Let suppose the triangle ABC as in figure attached:
Here we have cut off side AC at point D such that AD=AB and join BD.
As AD=AB, so ∠2=∠4
∠1>∠4 because Exterior angle is always greater than respective interior angle.
Similarly, ∠2>∠3.
Therefore ∠1>∠3 and BC>DC as size opposite to greater angle are also greater in length.
now, ⇒BC>DC
⇒BC>AC-AD as DC=AC-AD
⇒BC>AC-AB because AD=AB
Similarly, BC-AC<AB and BC-AB<AC
Hence it is proved that either side of triangle is always greater than difference of remaining two sides.
Hope my answer helps you.
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