In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one its altitudes.

Solution :

Let ABC be and equilateral triangle and let AD \(\perp\) BC.triangle

In \(\triangle\) ADB and ADC, we have :

AB = AC          (given)

AD = AD           (common side of triangle)

and \(\angle\) ADB = \(\angle\) ADB        (each 90)

By RHS criteria of congruence, we have :

\(\triangle\)  ADB \(\cong\) \(\triangle\) ADC

So, BD = DC or  BD = DC = \(1\over 2\) BC             ……..(1)

Since \(\triangle\)  ADB is a right triangle, angled at D, by Pythagoras theorem, we have :

\({AB}^2\) = \({AD}^2\) + \({BD}^2\)

\({AB}^2\) = \({AD}^2\) + \(({1\over 2}BC)^2\)        (from 1)

\({AB}^2\) = \({AD}^2\) + \({1\over 4}{BC}^2\)

\({AB}^2\) = \({AD}^2\) + \({AB}^2\over 4\)            (\(\therefore\)  BC = AB)

\({3\over 4}{AB}^2\) = \({AD}^2\)  or  \(3{AB}^2\) = \(4{AD}^2\)

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