state the prove pythagoras theoram
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In a right triangle, the square of the hypotenuse is equal to the sum of the square of the 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)
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
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
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