Question

prove that in a right angle triangle square of the hypotenuse is equal to sum of the square of other two sides

Answer 1

Here is the answer to your mate

 Given : A right ΔABC right angled at B

To prove : AC2 = AB2 + BC2

 Construction : Draw AD ⊥ AC

Proof : ΔABD and ΔABC

∠ADB = ∠ABC = 90°

∠BAD = ∠BAC  (common)

∴ ΔADB ∼ ΔABC  (by AA similarly criterion)

=>AD/AB=AB/AC

⇒ AD × AC = AB2    ...... (1)

 

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA  (common)

∴ ΔBDC ∼ ΔABC  (by AA similarly criterion)

=> CD/BC= BC/AC

⇒ CD × AC = BC2    ........ (2)

 
Adding (1) and (2) we get

AB2 + BC2 = AD × AC + CD × AC

= AC (AD + CD)

= AC × AC = AC2

∴ AC2 = AB2 + BC2

Answer 2

${ac}^{2} = {ab}^{2} + {bc}^{2}$
let ac 13
let ab 12
let bc 5

then

${ac}^{2} = {ab}^{2} + {bc}^{2}$
according to hypotenuse theorem
13
${13}^{2} = {12}^{2} + {5}^{2}$
169= 144+25
169=169

$\sqrt{169}$
13=13
Hence .