If D is the midpoint of the side BC of triangle ABC and AD is perpendicular to AC,then

3 b^{2} = a^{2} - c^{2}

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3 a^{2} = b^{2} - 3 c^{2}

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

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a^{2} + b^{2} = 5 c^{2}

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Solution

The correct option is B $3 b^{2} = a^{2} - c^{2}$

We know,

Using cosine rule

$c^{2} = b^{2} + a^{2} - 2 a b cos C = b^{2} + a^{2} - 2 a b (\frac{b}{\frac{a}{2}})$

$= b^{2} + a^{2} - {4 b}^{2}$

$∴ c^{2} = a^{2} - {3 b}^{2}$

$\Rightarrow {3 b}^{2} = a^{2} - c^{2}$

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Similar questions

In the adjoining figure, if BC = a, AC = b, AB = c and $∠ C A B = 120^{\circ}$ , then the correct relation is-

In the adjoining figure, if $B C = a, A C = b, A B = c$ and $∠ C A B = 120^{\circ}$ , then the correct relation is

In the adjoining figure, if $B C = a, A C = b, A B = c$ and $∠ C A B = 120^{\circ}$ ,
then which of the following is the correct relation?

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

△ A B C

cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

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