Question

In a obtuse triangle ABC AD is perpendicular to BC. Prove that AC^2=AB^2+BC^2-2BC*BD

Answer 1

o prove:  AC2  = AB2 + BC2  + 2 BC X BD

Proof: 
In ΔADC,

AC2  = AD2 + DC2  ( Pythagoras theorem)

AC2  = AD2 + (DB + BC)2 

AC2  = AD2 + DB2  + BC2  + 2DB X BC.

AC2  = AB2  + BC2  + 2 DB X BC. [ ∴ AD2 + DB2 = AB2  ]