In Figure, ABC is a triangle in which \(\angle\) ABC > 90° and AD \(\perp\) CB produced. Prove that \(AC^2= AB^2+ BC^2+ 2 BC.BD.\)

By applying Pythagoras Theorem in \(\triangle\) ADB, we get,
\(AB^2 = AD^2 + DB^2\) ……………………… (i)

Again, by applying Pythagoras Theorem in \(\triangle\) ACD, we get,
\(AC^2 = AD^2 + DC^2\)
\(AC^2 = AD^2 + (DB + BC)^2\)
\(AC^2 = AD^2 + DB^2 + BC^2 + 2DB × BC\)

From equation (i), we can write,
\(AC^2 = AB^2 + BC^2 + 2DB × BC\)
Hence, proved.