Question

IN AN EQUILATERAL TRIANGLE, PROVE THAT THREE TIMES THE SQUARE OF ONE SIDE IS EQUAL TO FOUR TIME'S THE SQUARE OF ONE OF ITS ALTITUDES. ​

Answer 1

Step-by-step explanation:

∆ ABC is a equilateral triangle

AB=BC=AC

AD is altitude

∆ABD is a right angled triangle

angle D=90°

AB^2=AD^2+BD^2

AB^2=AD^2+(AB/2)^2

AB^2= AD^2+AB^2/4

AB^2-AB^2/4=AD^2

4AB^2-AB^2= 4AD^2

3AB^2= 4AD^2