How to prove that difference of any two sides of a triangle?
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TO PROVE THAT: AC - AB < BC
CONSTRUCTION: From the longer side AC, cut a segment AD = AB
So, now, we need to prove that AC - AD < BC
=>TO PROVE first: that DC < BC
PROOF:< ABD = < ADB = a ( by construction)
< DBC = p, & < BDC = k
Since, k = a + A ( exterior < of a triangle) ……. (1)
p = a - C ( again by exterior < of a triangle) ….(2)
By comparing (1) & (2)
(a + A) > (a - C) , as all these angle variables >0
=> k > p
=> side opposite to k > side opposite to p
=> BC > DC
=> DC < BC
=> AC - AD < BC
=> AC - AB < BC
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