Question

Using the concept of slope, prove that medians of an equilateral triangle are perpendicular to the corresponding sides.

Answer 1

Let each side be $a㎝$

then $BD=DC=\frac{a}{2}$

($\because AD$ is median to BC)

We have altitude of equilateral $△ABC=\frac{\sqrt{3}}{2}a$

(If AD is also an altitude apart from being median)

Also $AD=\sqrt{{AC}^2-{DC}^2}=\sqrt{a^2-\frac{a^2}{4}}$

$=\sqrt{\frac{3a^2}{4}}=\frac{\sqrt{3}}{2}a$

$\therefore$Median of equilateral $△$ is also its altitude.