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| 1. |
To verify that the sum of any two sides of a triangles is always greater than the third side |
| Answer» Given ΔABC, extend BA to D such that AD = AC.Now in ΔDBC∠ADC = ∠ACD [Angles opposite to equal sides are equal]Hence ∠BCD > ∠ BDCThat is BD > BC [The side opposite to the larger (greater) angle is longer]Þ AB + AD > BC\xa0∴ AB + AC > BC [Since AD = AC)Thus sum of two sides of a triangle is always greater than third side. | |