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Inequality of a Triangle Theorem - III

Theorem: The sum of any two sides of a triangle is greater than the third side.
Given:
ΔABC.
To Prove:
(i) AB + AC > BC
(ii) AB + BC > AC
(iii) BC + AC > AB
Construction: Produce BA to D such that AC = AD. Join CD.

 

Proof:
In
ΔACD, we have
AC = AD ... [By construction]
...
D = 1 ... [Angles opposite to equal sides]
...
1 + 2 > D
...
BCD > D
Now in
ΔBCD, we have
         
BCD > BDC
...           BD > BC ... (side opposite to greater angle)
...    AB + AD > BC
...    AB + AC > BC ... [AC = AD]
Similarly, we can prove that
      AB + BC > AC and 
AC + BC > AB





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