Prove that the sum of any two sides of a triangle is greater than its third side
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Step-by-step explanation:
Construction: In ΔABC, extend AB to D in such a way that AD=AC.
In ΔDBC, as the angles opposite to equal sides are always equal, so,
∠ADC=∠ACD
Therefore,
∠BCD>∠BDC
As the sides opposite to the greater angle is longer, so,
BD>BC
AB+AD>BC
Since AD=AC, then,
AB+AC>BC
Hence, sum of two sides of a triangle is always greater than the third side.
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