Proof Pythagoras theorem i.e. $c^{2} = a^{2} + b^{2}$

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Solution

Following is the proof

Consider a right angled triangle ABC which is right angled at B. Let AB=a, BC=b and AC=c
From B , draw a line perpendicular to side AC, which meets AC at D.
Now, consider triangles ADB and ABC
$∠$ A = $∠$ A
$∠$ ADB = $∠$ ABC = 90 $^{\circ}$
Therefore, by AA similarity criterion, triangles ADB and ABC are similar.
Therefore $(\frac{A D}{A B}) = (\frac{A B}{A C})$
$A B^{2} = A D \times A C$ ...............(1)
Similarly $(\frac{B C}{C D}) = (\frac{A C}{B C})$
This implies, $B C^{2} = A C \times C D$ ...............(2)
Adding 1 and 2
$(A B^{2}) + (B C^{2}) = (A D \times A C) + (A C \times C D) = A C (A D + C D) = A C \times A C = A C^{2}$
Thus, $a^{2} + b^{2} = c^{2}$

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|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

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