Explanation:
Draw AE⊥BC
In △AEB and △AEC, we have
AB=AC
AE=AE [common]
and, ∠b=∠c [because AB=AC]
∴ △AEB≅△AEC
⇒ BE=CE
Since △AED and △ABE are right-angled triangles at E.
Therefore,
AD2=AE2+DE2 and AB2=AE2+BE2
⇒ AB2−AD2=BE2−DE2
⇒ AB2−AD2=(BE+DE)(BE−DE)
⇒ AB2−AD2=(CE+DE)(BE−DE) [∵BE=CE]
⇒ AB2−AD2=CD.BD
AB2−AD2=BD.CD
