If D is the mid-point of side BC of a 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 A $3 b^{2} = a^{2} - c^{2}$

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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?

△ 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}

In a triangle ABC, $2 c a s i n \frac{A - B + C}{2}$ is equal to

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