In an equilateral triangle ABC, BE is perpendicular to side CA. Prove that :

AB² + BC² + CA² = 4BE²

Given: ABC is an equilateral triangle and BE ⊥ AC.

To prove: AB² + BC² + CA² = 4BE²

Proof: Since △ABC is equilateral, we know that: AB = BC = CA.

Since BE ⊥ AC and the perpendicular from a vertex in an equilateral triangle bisects opposite side, so, E is the mid-point of AC.

AE = EC = $\dfrac{\text{AC}}{2}$ = $\dfrac{\text{BC}}{2}$

In Δ BEC, using Pythagoras theorem,

BC² = BE² + CE²

⇒ BC² = BE² + $\Big(\dfrac{\text{BC}}{2}\Big)$²

⇒ BE² = BC² - $\Big(\dfrac{\text{BC}^2}{4}\Big)$

⇒ BE² = $\Big(\dfrac{\text{4BC}^2 - \text{BC}^2}{4}\Big)$

⇒ BE² = $\Big(\dfrac{\text{3BC}^2}{4}\Big)$

⇒ 4BE² = 3BC²

Since AB = BC = CA, we have:

⇒ AC² = AB² = BC²

⇒ 4BE² = AC² + AB² + BC²

Hence, AB² + BC² + CA² = 4BE².