Question

In an equilateral triangle $\Delta ABC$ if $AN$ perpendicular to $BC$, prove that $A{N}^2=3B{N}^2$

Answer 1

If $AN\perp BC$

Then $△ANB$ will be a right angle $△$ ,

Using Pythagorean theorem, we get

$(AN{)}^2+(BN{)}^2=(AB{)}^2$

On subtracting $(BN{)}^2$ from both side

$(AN{)}^2=(AB{)}^2-(BN{)}^2$

Since triangle $ABC$ is equilateral, perpendicular segment $ANtoBC$ bisects line segment $BC$,

Therefore, $AB=2BN$ then

$(AN{)}^2=(2BN{)}^2-(BN{)}^2$

$(AN{)}^2=4(BN{)}^2-(BN{)}^2$

$(AN{)}^2=3(BN{)}^2$

Hence proved.