Question

prove that the sum of any two side of the 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.
hope it helps

Answer 2

means Pythagoras theorem we have to prove ....

let abc is a triangle in which angle bac = 90 bc is hypotenuse
to prove ab^2+ac^2=bc^2
con
draw ad perpendicular bc
angle adb= angle adc =90
now these pic will help u these pic is a proof