The perpendicular from A on side …

Given: The perpendicular from A on side BC of an{tex}\vartriangle {/tex} ABC intersect BC at Such that BC = 3CD

To prove:2AB² = 2AC² + BC² 
Proof:In right triangle ADB,
AB² = AD² + BC²  (1).......[By Pythagoras theorem]
In right triangle ADC,
AC² = AD² + CD² (2).....[By Pythagoras theorem]
Subtracting (2) from (1), we get
AB² - AC² = BD² - CD²
= (BD + CD)(BD - CD)
= (BC)(3CD - CD) ........ {tex}\because {/tex} BD = 3CD(given)
= (BC)(2CD) = 2(BC)(CD)
=2(BC){tex}\left( {\frac{1}{4}BC} \right){/tex}
DB = 3CD
{tex} \Rightarrow \frac{{BD}}{{CD}} = 3{/tex}
{tex} \Rightarrow \frac{{DB}}{{CD}} + 1{/tex}
= 3 + 1
{tex} \Rightarrow \frac{{DB + CD}}{{CD}}=4{/tex}
{tex}\Rightarrow \frac{{BC}}{{CD}} = 4{/tex}
{tex}\Rightarrow CD = \frac{1}{4}BC = B{C^2}{/tex}
{tex}\Rightarrow {/tex} 2(AB² - AC²) = BC²
{tex}\Rightarrow {/tex} 2AB² - 2AC² = BC²
{tex}\Rightarrow {/tex} 2AB² = 2AC² + BC²