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proofs of pythagoras theorem

Asked by squishing_gazal...(student) , on 9/6/12


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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 × AC = AB2 ...... (1)

 

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA (common)

∴ ΔBDC ∼ ΔABC (by AA similarly criterion)

⇒ 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

Posted by Anuvinda Murali(student)on 9/6/12

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