Prove that the sum of two sides of a triangle is alwayes greater than

1.	Prove that the sum of two sides of a triangle is alwayes greater than the third angle
Answer» Answer: 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. (Mark me as the brainliest)

Prove that the sum of two sides of a triangle is alwayes greater than the third angle

Answer»

Answer:

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.

(Mark me as the brainliest)

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Prove that the sum of two sides of a triangle is alwayes greater than the third angle

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