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Question
Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
Solution
Given: A right ΔABC right angled at B
To prove : AC2 = AB2 + BC2
Construction: Draw AD ⊥ AC
Proof: ΔADB 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
Hence Proved.
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