Question

In an isosceles triangle AB =AC. BD is perpendicular to AC prove that BD squared - CD square is equal to two times CD into AD

Answer 1

Given: AB=AC , BD perpendicular AC

To Prove: BD^2 -DC^2 = 2DC* AD

Proof: BC^2 =CD^2 + BD^2 [PYTHA. THEO.]

=> BD^2 =BC^2-CD^2

AB^2= AD^2+ BD^2 [PYTHA. THEO.]

=> BD^2 =AB^2-AD^2

=[ AB+AD ] * [ AB-AD ] {A^2-B^2= [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+DC^2+AD*DC

BD^2 =2AD*DC+DC^2

BD^2 -DC^2 = 2DC* AD