Question

AD is perpendicular to the side BC of an equilateral ΔABC. Prove that 4AD² = 3AB².

Answer 1

In equilateral triangle AD ⊥ BC.

⇒ BD = DC = `"BC"/(2)`   ...(In equilateral triangle altitude bisects the opposite side)

In right triangle ABD,
AB² = AD² + BD²

= `"AD"^2 + ("BC"/2)^2`

= `(4"AD"^2 + "BC"^2)/(4)`

= `(4"AD"^2 + "BC"^2)/(4)`  ...(Since AB = BC)

⇒ 4AB² = 4AD² + AB²
⇒ 3AB² = 4AD²
Hence proved..