In the figure, given below, $A D ⊥ B C$ .
Prove that : $c^{2} = a^{2} + b^{2} - 2 a x$ .

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Solution

For the right angled $△ A D B$ , as per the pythagoras theorem, ${A B}^{2} = {A D}^{2} + {B D}^{2}$
$=> {(c)}^{2} = {(h)}^{2} + {(a - x)}^{2}$
$=> c^{2} = h^{2} + a^{2} + x^{2} - 2 a x$
$=> h^{2} = c^{2} - a^{2} - x^{2} + 2 a x$ -- (1)
Similarly,
For the right angled $△ A C D$ , as per the pythagoras theorem, ${A C}^{2} = {A D}^{2} + {C D}^{2}$
$=> b^{2} = {(h)}^{2} + {(x)}^{2}$ -- (2)
From, eq.(1) and eq.(2)
$=> b^{2} = c^{2} - a^{2} - x^{2} + 2 a x + x^{2}$
$=> b^{2} = c^{2} - a^{2} + 2 a x$
$=> c^{2} = a^{2} + b^{2} - 2 a x$

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⊥

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