Question

In an equilateral $△$ ABC, AD $\perp$ BC. prove that ${AD}^2=3{BD}^2$

Answer 1

The altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.

$AC=AB=a$

$CD=DB=\frac{a}{2}$

$(DB{)}^2=\frac{a^2}{4}$

In right angled triangle $△ADB$

$(AB{)}^2=(AD{)}^2+(DB{)}^2$

$(a{)}^2=(AD{)}^2+\frac{(a{)}^2}{4}$

$A{D}^2=a^2-\frac{a^2}{4}$

$A{D}^2=\frac{3a^2}{4}=3\frac{a^2}{4}=3B{D}^2$

$A{D}^2=3B{D}^2$

Hence proved.