Answer:

let BD=3x and CD=x

in △ABD. 

by Pythagoras

AB2=BD2+AD2

AD2=AB2−BD2......let this eq 1

Now in △ACD.

by Pythagoras

AC2=AD2+CD2

AD2=AC2−CD2....let this eq 2

from 1 and 2

we get

AB2−BD2=AC2−CD2

AB2−AC2=BD2−CD2

AB2−AC2=9x2−x2

AB2−AC2=8x2

AB2−AC2=2(4x2)

AB2−AC2=2(BC/2)2

AB