# In the given figure, $AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that(i)$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^{2}$(ii) $\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^2$(iii) $\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$"

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Given:

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

To do:

We have to prove that

(i) $\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \times \mathrm{DM}+(\frac{\mathrm{BC}}{2})^{2}$

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

(iii) $\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$

Solution:

(i) In $\triangle \mathrm{AMC}$,

$\angle \mathrm{AMC}=90^{\circ}$

This implies, by Pythagoras theorem,

$\mathrm{AC}^{2}=\mathrm{AM}^{2}+\mathrm{MC}^{2}$

$\mathrm{AC}^{2}=\mathrm{AM}^{2}+(\mathrm{MD}^{2}+\mathrm{DC}^{2})$         ($\mathrm{MC}=\mathrm{MD}+\mathrm{DC}$)

$\mathrm{AC}^{2}=\mathrm{AM}^{2}+(\mathrm{MD}+\frac{1}{2} \mathrm{BC})^{2}$

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

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

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

Hence proved.

(ii) 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.

(iii) 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.

Updated on 10-Oct-2022 13:21:47