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In the given figure, $AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that
$\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$
"
Given:
$AD$ is a median of a triangle $ABC$ and $AM \perp BC$.
To do:
We have to prove that $\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$
Solution:
In $\triangle \mathrm{AMC}$,
$\angle \mathrm{AMC}=90^{\circ}$
This implies, by Pythagoras theorem,
$\mathrm{AC}^{2}=\mathrm{AM}^{2}+\mathrm{MC}^{2}$
$\mathrm{AC}^{2}=\mathrm{AM}^{2}+(\mathrm{MD}^{2}+\mathrm{DC}^{2})$ ($\mathrm{MC}=\mathrm{MD}+\mathrm{DC}$)
$\mathrm{AC}^{2}=\mathrm{AM}^{2}+(\mathrm{MD}+\frac{1}{2} \mathrm{BC})^{2}$
$\mathrm{AC}^{2}=\mathrm{AM}^{2}+\mathrm{MD}^{2}+(\frac{\mathrm{BC}}{2})^{2}+2 \mathrm{MD} \times \frac{\mathrm{BC}}{2}$
$\mathrm{AC}^{2}=(\mathrm{AM}^{2}+\mathrm{MD}^{2})+\mathrm{MD} \times \mathrm{BC}+(\frac{\mathrm{BC}}{2})^{2}$
$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{MD} \times \mathrm{BC}+(\frac{\mathrm{BC}}{2})^{2}$......(i)
In $\triangle \mathrm{AMB}$, by Pythagoras theorem,
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{BM}^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+(\mathrm{BD}-\mathrm{MD})^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+(\frac{\mathrm{BC}}{2}-\mathrm{MD})^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{MD}^{2}+(\frac{\mathrm{BC}}{2})^{2}-\frac{2 \mathrm{BC}}{2} \times \mathrm{MD}$
$\mathrm{AB}^{2}=(\mathrm{AM}^{2}+\mathrm{MD}^{2})+(\frac{\mathrm{BC}}{2})^{2}-\mathrm{BC} \times \mathrm{MD}$
$\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{MD}+(\frac{\mathrm{BC}}{2})^{2}$.........(ii)
Adding equations (i) and (ii), we get,
$AC^{2}+AB^{2}=2AD^{2}+\frac{2 BC^{2}}{4}$
$AC^{2}+AB^{2}=2AD^{2}+\frac{BC^{2}}{2}$
Hence proved.