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In the given figure, $AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that
$\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^2$
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Given:

$AD$ is a median of a triangle $ABC$ and $AM \perp BC$.

To do:

We have to prove that $\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^2$

Solution:

In $\triangle \mathrm{AMB}$, by Pythagoras theorem,

$\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{BM}^{2}$

$\mathrm{AB}^{2}=\mathrm{AM}^{2}+(\mathrm{BD}-\mathrm{MD})^{2}$

$\mathrm{AB}^{2}=\mathrm{AM}^{2}+(\frac{\mathrm{BC}}{2}-\mathrm{MD})^{2}$

$\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{MD}^{2}+(\frac{\mathrm{BC}}{2})^{2}-\frac{2 \mathrm{BC}}{2} \times \mathrm{MD}$

$\mathrm{AB}^{2}=(\mathrm{AM}^{2}+\mathrm{MD}^{2})+(\frac{\mathrm{BC}}{2})^{2}-\mathrm{BC} \times \mathrm{MD}$

$\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{MD}+(\frac{\mathrm{BC}}{2})^{2}$

Hence proved.

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Updated on: 10-Oct-2022

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