Question

Show that the sum of any two sides of a triangle is greater than the third side​

Answer 1

ANSWER

Construction: In ΔABC, extend AB to D in such a way that AD=AC.

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,

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

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=∠ACDTherefore,

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=∠ACDTherefore,∠BCD>∠BDC

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=∠ACDTherefore,∠BCD>∠BDCAs the sides opposite to the greater angle is longer, so,

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=∠ACDTherefore,∠BCD>∠BDCAs the sides opposite to the greater angle is longer, so,BD>BC

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=∠ACDTherefore,∠BCD>∠BDCAs the sides opposite to the greater angle is longer, so,BD>BCAB+AD>BC

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=∠ACDTherefore,∠BCD>∠BDCAs the sides opposite to the greater angle is longer, so,BD>BCAB+AD>BCSince AD=AC, then,

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=∠ACDTherefore,∠BCD>∠BDCAs the sides opposite to the greater angle is longer, so,BD>BCAB+AD>BCSince AD=AC, then,AB+AC>BC

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=∠ACDTherefore,∠BCD>∠BDCAs the sides opposite to the greater angle is longer, so,BD>BCAB+AD>BCSince AD=AC, then,AB+AC>BCHence, sum of two sides of a triangle is always greater than the third side.

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Answer 2

Answer:

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Step-by-step explanation:

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