In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1 : 3.

Prove that : 2AC² = 2AB² + BC².

Given,

D divides BC in the ratio 1 : 3.

⇒ BD : DC = 1 : 3

Let BD = x and DC = 3x

From figure,

⇒ BC = BD + DC = x + 3x = 4x.

$\Rightarrow \dfrac{BD}{BC} = \dfrac{x}{4x} \\[1em] \Rightarrow \dfrac{BD}{BC} = \dfrac{1}{4} \\[1em] \Rightarrow BD = \dfrac{1}{4}BC. \\[1em] \Rightarrow \dfrac{CD}{BC} = \dfrac{3x}{4x} \\[1em] \Rightarrow \dfrac{CD}{BC} = \dfrac{3}{4} \\[1em] \Rightarrow CD = \dfrac{3}{4}BC.$

In right angle triangle ADC,

By pythagoras theorem,

⇒ AC² = AD² + CD² …………(1)

In right angle triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD² …………(2)

Subtracting equation (2) from (1), we get :

⇒ AC² - AB² = AD² + CD² - (AD² + BD²)

⇒ AC² - AB² = CD² - BD²

Substituting value of BD and CD in above equation, we get :

$\Rightarrow AC^2 - AB^2 = \Big(\dfrac{3}{4}BC\Big)^2 - \Big(\dfrac{1}{4}BC\Big)^2 \\[1em] \Rightarrow AC^2 - AB^2 = \dfrac{9}{16}BC^2 - \dfrac{1}{16}BC^2 \\[1em] \Rightarrow AC^2 - AB^2 = \dfrac{8}{16}BC^2 \\[1em] \Rightarrow AC^2 - AB^2 = \dfrac{1}{2}BC^2 \\[1em] \Rightarrow 2(AC^2 - AB^2) = BC^2 \\[1em] \Rightarrow 2AC^2 - 2AB^2 = BC^2 \\[1em] \Rightarrow 2AC^2 = 2AB^2 + BC^2.$

Hence, proved that 2AC² = 2AB² + BC².