Question

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

Prove that : 2AC² = 2AB² + BC²

Answer 1

Here,
BD : DC = 1 : 3.
⇒ BD = `1/4"BC" and CD = 3/4`BC

AC² = AD² + CD² and AB² = AD² + BD² 
Therefore,
AC² - AB² = CD² - BD²

= `( 3/4"BC" )^2 - ( 1/4 "BC" )^2 `

= `9/16 "BC"^2 -  1/16 "BC"^2`

= `1/2"BC"^2`

∴ 2AC² - 2AB² = BC²
2AC² = 2AB² + BC²
Hence proved.