# How we will prove that the sum of the measure of any two sides of a triangleis greater than the third side

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by marini

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by marini

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We can extend BA past A into a straight line.

There exists a point D such that DA = CA.

Therefore, \angle ADC = \angle ACD because isosceles triangle have two equal angles.

Thus, \angle BCD > \angle BDC by Euclid's fifth common notion.

Since \triangle DCB is a triangle having \angle BCD greater than \angle BDC, this means that BD > BC.

But BD = BA + AD, and AD = AC.

Thus, BA + AC > BC.

A similar argument shows that AC + BC > BA and BA + BC > AC.

\blacksquare