Question

In a `DeltaABC`, right angled at `A`, a point `D` is on side `AB`. Prove that `CD^2=BC^2+BD^2`?

Answer 1

`CD^2!=BC^2+BD^2`, but

`CD^2=BC^2+BD^2-2BDxxAB`

Explanation:

With a point `D` on `AB`, we have the figure as

In the right angled triangle `ABC` right angled at `A`, we have

`BC^2=AB^2+AC^2=(AD+BD)^2+AC^2`

= `AD^2+BD^2+2xxBDxxAD+AC^2`

or `BC^2+BD^2`

= `AD^2+AC^2+2BD^2+2xxBDxxAD`

= `CD^2+2BD(BD+AD)`

= `CD^2+2BDxxAB`

or `CD^2=BC^2+BD^2-2BDxxAB`