# 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.

In $$\triangle$$ ADB and ADC, we have :

AB = AC          (given)

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  $$3{AB}^2$$ = $$4{AD}^2$$