Question

in a quardilateral triangle ABCD ,D is a point on side BC such that BD = 1/3BC. prove that9AD^=7AB^

Answer 1

BD= 1/3BC
therefore DC=2/3 BC
construct AE perpendicular to BC so BE =1/2 BC
in rt triangle AED,  ADsq = AEsq +DEsq             eq no 1
in rt triangle AEB,  ABsq = AEsq + BEsq             eq no 2
sbtract 2 from 1
ADsq - ABsq =DEsq -BEsq
ADsq = ABsq + [BE - BD] sq - BE sq
ADsq = ABsq +[BC/2 - BC/3] sq - [BC/2] sq
          =AB sq + [3BC - 2BC/6] sq -BC sq /4
          =AB sq + BCsq/36 -BCsq/4
          =36AB sq + AB sq - 9 AB sq/36
          =28AB sq/36
ADsq =7AB sq/9
therefore      9AD sq= 7AB sq