Question

theorem 7.8 sum of two sides of a triangle is greater than the third side

Answer 1

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

Answer 2

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 > ∠ BDC That is BD > BC [The side opposite to the larger (greater) angle is longer] Þ AB + AD > BC  ∴ AB + AC > BC [Since AD = AC) Thus sum of two sides of a triangle is always greater than the third side, hence proved