# The perpendicular from A on side BC of a $∆ABC$ intersects BC at D such that $DB = 3CD$ (see the figure). Prove that $2AB^2 = 2AC^2 + BC^2$."

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Given:

The perpendicular from A on side BC of a $∆ABC$ intersects BC at D such that $DB = 3CD$.

To do:

We have to prove that $2AB^2 = 2AC^2 + BC^2$.

Solution:

In $\triangle \mathrm{ADC}$,

$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}$

$DB=3CD$

$BC=DC+BD$

$BC=DC+3CD$

$BC=4CD$

$CD=\frac{1}{4}(BC)$

In $\triangle \mathrm{ADB}$,

$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$

$=\mathrm{AD}^{2}+(\mathrm{BC}-\mathrm{DC})^{2}$

$=\mathrm{AD}^{2}+\mathrm{BC}^{2}+\mathrm{DC}^{2}-2 \mathrm{BC} . \mathrm{DC}$

$=(\mathrm{AD}^{2}+\mathrm{DC}^{2})+\mathrm{BC}^{2}-2 \mathrm{BC}.\mathrm{DC}$

$=\mathrm{AC}^{2}+\mathrm{BC}^{2}-2 \mathrm{BC}.\frac{1}{4} \mathrm{BC}$

$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}-\frac{1}{2} \mathrm{BC}^{2}$

$\mathrm{AB}^{2} = \mathrm{AC}^{2}+\frac{1}{2}\mathrm{BC}^{2}$

$2\mathrm{AB}^{2}=2 \mathrm{AC}^{2}+\mathrm{BC}^{2}$

Hence proved.

Updated on 10-Oct-2022 13:21:36