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In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
To do:
We have to prove that three times the square of one side of an equilateral triangle is equal to four times the square of one of its altitudes.
Solution:
Let in a $\triangle \mathrm{ABC}$,
$\mathrm{AB}=\mathrm{BC}=\mathrm{AC}$
$\mathrm{AD} \perp \mathrm{BC}$
In $\triangle \mathrm{ADB}$,
$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$
$\mathrm{AD}^{2}=\mathrm{AB}^{2}-\mathrm{BD}^{2}$
$=\mathrm{AB}^{2}-(\frac{1}{2} \mathrm{BC})^{2}$
$=\mathrm{AB}^{2}-\frac{1}{4} \mathrm{BC}^{2}$
$4 \mathrm{AD}^{2}=4 \mathrm{AB}^{2}-\mathrm{BC}^{2}$
$=4 \mathrm{AB}^{2}-\mathrm{AB}^{2}$ (Since $BC=AB$)
$4 \mathrm{AD}^{2} =3 \mathrm{AB}^{2}$
Hence proved.
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