prove that in a right angle triangle the square of the hypotenuse is equal to the sum of the square of remaining two side
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The Pythagoras Theorem. Let's have a look at what Mr Pythagoras stated when he came up with the Theorem, Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of theremaining two sides. ... Thehypotenuse is the longest side of theright-angled triangle.
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Answer:
Given :
A right triangle ABC right angled at B.
To prove :
AC² = AB² + BC²
Construction :
Draw BD ⊥ AC
Proof :
In Δ ADB and Δ ABC
∠ A = ∠ A [ Common angle ]
∠ ADB = ∠ ABC [ Both are 90° ]
∴ Δ ADB Similar to Δ ABC [ By AA similarity ]
So , AD / AB = AB / AC [ Sides are proportional ]
= > AB² = AD . AC ... ( i )
Now in Δ BDC and Δ ABC
∠ C = ∠ C [ Common angle ]
∠ BDC = ∠ ABC [ Both are 90° ]
∴ Δ BDC Similar to Δ ABC [ By AA similarity ]
So , CD / BC = BC / AC
= > BC² = CD . AC ... ( ii )
Now adding both equation :
AB² + BC² = CD . AC + AD . AC
AB² + BC² = AC ( CD + AD )
AB² + BC² = AC² .
AC² = AB² + BC² .
Hence proved .
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