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

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

We can write it as;
\( AD^2 = AB^2 - DB^2 \)……………….. (i)

By applying Pythagoras Theorem in \(\triangle\) ADC, we get,
\(AD^2 + DC^2 = AC^2\)

From equation (i),
\(AB^2 - BD^2 + DC^2 = AC^2\)
\(AB^2 - BD^2 + (BC - BD)^2 = AC^2\)
\(AC^2 = AB^2 - BD^2 + BC^2 + BD^2 - 2BC × BD\)
\(AC^2= AB^2 + BC^2 - 2BC × BD\)
Hence, proved.