In the given figure, $∠ B < 90$ and segment $A D ⊥ B C$ , show that
(i) $b^{2} = h^{2} + a^{2} + x^{2} - 2 a x$
(ii) $b^{2} = a^{2} + c^{2} - 2 a x$

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Solution

As $△ A D C$ is right angled at D

So applying Pythagoras theorem states that $A C^{2} = A D^{2} + D C^{2}$
$b^{2} = h^{2} + (a - x)^{2}$
$b^{2} = h^{2} + a^{2} + x^{2} - 2 a x$ (1)
Hence (i) proved.

Also, $△ A D B$ is right-angled at D.

So applying Pythagoras theorem states that $c^{2} = h^{2} + x^{2}$ (2)
Substituting (2) in (1)
$b^{2} = h^{2} + x^{2} + a^{2} - 2 a x$

$b^{2} = (h^{2} + x^{2}) + a^{2} - 2 a x$

$b^{2} = c^{2} + a^{2} - 2 a x$

$b^{2} = a^{2} + c^{2} - 2 a x$
Hence (ii) proved.

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In the given figure, ∠B<90 and segment AD ⊥ BC, show that(i) b2=h2+a2+x2−2ax(ii) b2=a2+c2−2ax

In the given figure, $∠ B < 90$ and segment $A D ⊥ B C$ , show that
(i) $b^{2} = h^{2} + a^{2} + x^{2} - 2 a x$
(ii) $b^{2} = a^{2} + c^{2} - 2 a x$