Question

In an isosceles triangle ABC with AB = AC and BD is perpendicular to AC. Prove that
BD^2 - CD^2 = 2CD × AD
VERY URGENT PLZZ

Answer 1

Given: AB=AC , BD _|_ AC

To Prove: BD2 -DC2 = 2DC* AD

Proof: BC2 =CD2 + BD2     [PYTHA. THEO.]

      => BD2 =BC2-CD2

           AB2 = AD2+ BD2       [PYTHA. THEO.]

     => BD2 =AB2-AD2

                   =[ AB+AD ] * [ AB-AD ]     {A2-B2 = [A+B] [A-B] }

                   =[ AC+AD ] [ AC-AD]          { Since AB=AC}

                   =[ AC+AD ] * DC

                   =AC*DC + AD*DC

                  =[AD+DC]*DC + AD*DC  

                  =AD*DC+DC2+AD*DC  

       BD2   =2AD*DC+DC2

       BD2 -DC2 = 2DC* AD
=> BD²-DC² = 2CD×AF