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Dear Student,

Statement : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides.
 

 

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)

Now, corresponding sides of similar triangles are proportional, 

    

⇒ AD × AC = AB²    ...... (1)

 

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA  (common)

∴ ΔBDC ∼ ΔABC  (by AA similarly criterion)

Now, corresponding sides of similar triangles are proportional, 

⇒ 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 

Regards