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