Let $A D$ be a median of the $△ A B C$ . If $A E$ and $A F$ are medians of the triangle $A B D$ and $A D C$ , respectively, and $A D = m_{1}, A E = m_{2}, A F = m_{3}$ , then $\frac{a^{2}}{8}$ is equal to

m_{2}^{2} + m_{3}^{2} - 2 m_{1}^{2}

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m_{1}^{2} + m_{2}^{2} - 2 m_{3}^{2}

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m_{1}^{2} + m_{3}^{2} - 2 m_{2}^{2}

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m_{1}^{2} + m_{2}^{2} - 2 m_{3}^{2}

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Solution

The correct option is A $m_{2}^{2} + m_{3}^{2} - 2 m_{1}^{2}$

In $△ A B C, A D^{2} = m_{1}^{2} = \frac{c^{2} + b^{2}}{2} - \frac{a^{2}}{4}$
In $△ A B D, A E^{2} = m_{2}^{2} = \frac{A D^{2} + c^{2}}{2} - \frac{{(\frac{a}{2})}^{2}}{4}$
In $△ A D C, A F^{2} = m_{3}^{2} = \frac{A D^{2} + b^{2}}{2} - \frac{{(\frac{a}{2})}^{2}}{4}$
$∴ m_{2}^{2} + m_{3}^{2} = A D^{2} + \frac{b^{2} + c^{2}}{2} - \frac{a^{2}}{8} = m_{1}^{2} + m_{1}^{2} + \frac{a^{2}}{4} - \frac{a^{2}}{8} = 2 m_{1}^{2} + \frac{a^{2}}{8}$
Or,
$m_{2}^{2} + m_{3}^{2} - 2 m_{1}^{2} = \frac{a^{2}}{8}$

Suggest Corrections

Similar questions

Q. Let AD be a median of the

△ A B C .

If AE and AF are medians of the triangle ABD and ADC, respectively, and

B D = \frac{a}{2}, A D = m_{1}, A E = m_{2}, A F = m_{3},

then

\frac{a^{2}}{8}

is equal to:

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