In the given fig; AD is perpendicular to BC produced. Prove that :
$c^{2} = a^{2} + b^{2} + 2 a x$

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Solution

In right-angle $Δ A B D$ , from Pythagorus Theorem

$A B^{2} = A D^{2} + B D^{2}$

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

$c^{2} = h^{2} + a^{2} + x^{2} + 2 a x$ ................. (1)

In right-angle $Δ A C D$ , from Pythagorus Theorem

$A C^{2} = A D^{2} + C D^{2}$

$b^{2} = h^{2} + x^{2}$

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

Substitute the above value of $h^{2}$ in equation (1),

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

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

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Similar questions

A D ⊥ B C

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

b^{2} = h^{2} + a^{2} + x^{2} - 2 a x

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

In the given figure, D is the midpoint of side BC and AE $⊥$ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that

(i) $b^{2} = p^{2} + a x + \frac{a^{2}}{4}$

(ii) $c^{2} = p^{2} - a x + \frac{a^{2}}{4}$

(iii) $(b^{2} + c^{2}) = 2 p^{2} + \frac{1}{2} a^{2}$

(iv) $(b^{2} - c^{2}) = 2 a x$

In Fig. 7.223, D is the mid-point of side BC and $A E ⊥ B C$ . If BC = a AC=b, AB=c, ED=z, AD = p and AE =h, prove that:

(i) $b^{2} = p^{2} + a x + \frac{a^{2}}{4}$
(ii) $c^{2} = p^{2} + a x + \frac{a^{2}}{4}$
(iii) $b^{2} + c^{2} = 2 p^{2} + \frac{a^{2}}{4}$

∠ B

⊥

\begin{matrix} i) b^{2} = h^{2} + a^{2} + x^{2} - 2 a x i i) b^{2} = a^{2} + c^{2} - 2 a x \end{matrix}

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