Question

Prove that the angle of any two side of a triangle is greater than its third side

Answer 1

Sorry, there is no figure:

But draw a right angled triangle at C with upper angle as D and the lower as B Join CA where A is on hypotenuse(Middle of hypotenuse).

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:

∠ABD>∠CDB. Hence we have AD>AB( because the side opposite to a larger angle is longer). AD = AC+BC. Hence the sum of two sides of a triangle is larger than the third side.