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Prove that the sum of two sides of any triangle is greater than the third side .

Or, Prove that the diagonals of a parallelogram bisect each other.


We can extend BA past A into a straight line.There exists a point D such that DA=CA.Therefore, ∠ADC=∠ACD because isosceles triangle have two equal angles.Thus, ∠BCD>∠BDC by Euclid's fifth common notion.Since △DCB is a triangle having ∠BCD greater than ∠BDCthis 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.
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