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Inequality of a Triangle Theorem - II

Theorem 2: In a triangle, the greater angle has the longer side opposite to it.
Given: In
ΔABC, such that B > C
To Prove: AC > AB

Proof:
In
ΔABC, there may be three possibilities and only one should be true.
(i) AB = AC (ii) AC < AB or (iii) AC > AB

Case I: If AC = AB, B = C, which is contrary to the given condition (B > C). 
AC
AB

Case II: If AC < AB, since the longer side has the greater angle opposite to it, 

C > B
This is again contrary to what is given

 AC AB

Case III: Now we are left with the only possibility namely AC > AB, which must be true.

AC > AB

 





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