Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.

Since, ABC is an equilateral triangle. Let each side be x cm.

PN, PM, and PL are perpendicular on side AB, AC and BC respectively. AD is any altitude from point A on side BC.

Join PA, PB and PC.

Area of ∆ABC = $\dfrac{1}{2}$ × Base × Altitude

= $\dfrac{1}{2} × BC × AD = \dfrac{x}{2}AD$……..(1)

Area of ∆APB = $\dfrac{1}{2}$ × AB × NP ……..(2)

Area of ∆APC = $\dfrac{1}{2}$ × AC × MP ……..(3)

Area of ∆BPC = $\dfrac{1}{2}$ × BC × LP ……..(4)

Adding (2), (3) and (4)

⇒ Area of (∆APB + ∆APC + ∆BPC) = $\dfrac{1}{2}$ × (AB × NP + AC × MP + BC × LP)

From figure,

$⇒ \text{Area of ∆ABC} = \dfrac{1}{2} [\text{AB} \times \text{NP} + \text{AC} \times \text{MP} + \text{BC} \times \text{LP}] \\[1em] = \dfrac{1}{2} [x \times \text{NP} + x \times \text{MP} + x \times \text{LP}] \\[1em] = \dfrac{x}{2} [\text{NP} + \text{MP} + \text{LP}]$

∴ Area of ∆ABC = $\dfrac{x}{2}$ [NP + MP + LP] …….(5)

From (1) and (5),

$\dfrac{x}{2}$ × AD = $\dfrac{x}{2}$ × (NP + LP + MP)

⇒ AD = NP + LP + MP.

Hence, proved that the sum of three perpendiculars is equal to the altitude of the triangle.