Answer:

Draw AE⊥BC, meeting BC at D.

Applying Pythagoras theorem in right-angled triangle AED,

ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.

BE = CE

DE + CE = DE + BE = BD

????????² = ????????² + ????????²

????????² = ????????² − ????????² …(i)

In ∆ACE,

????????² = ????????² + ????????²

????????² = ????????² − ????????² … (????????)

Using (i) and (ii),

????????² − ????????² = ????????² − ????????²

????????² − ????????² = ????????² − ????????²

=> (DE + CE) (DE – CE)

=> (DE + BE) CD

=> BD.CD