IN AN EQUILATERAL TRIANGLE, PROVE THAT THE THREE TIMES THE SQUARE OF ONE SIDE IS EQUAL TO FOUR TIMES THE SQURAE OF ONE OF ITS ALTITUDE.
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LET ABC BE AN EQUILATERAL TRIANGLE AND LET AD IS PERPENDICULAR TO DC
NOW, IN RIGHT ANGLE TRIANGLE ADB, WE HAVE
AB ^2 = AD^2 + BD ^2
=> AB ^2 = AD^2 + (1/2 BC^2)
=> AB^2 = AD^2 + 1/4BC^2
=> AB^2 = AD^2 + AB^2/ 4. (AB=BC)
=> AB^2 - AB^2/4 = AD^2
=> 3AB^2 / 4 = AD^2
=> 3AB^2 = 4AD^2
NOW, IN RIGHT ANGLE TRIANGLE ADB, WE HAVE
AB ^2 = AD^2 + BD ^2
=> AB ^2 = AD^2 + (1/2 BC^2)
=> AB^2 = AD^2 + 1/4BC^2
=> AB^2 = AD^2 + AB^2/ 4. (AB=BC)
=> AB^2 - AB^2/4 = AD^2
=> 3AB^2 / 4 = AD^2
=> 3AB^2 = 4AD^2
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