Question

In any∆ABC prove that: OA+OB<AB​

Answer 1

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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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