In fig., ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC.BD.

To Prove: AC² = AB² + BC² + 2BC.BD
Proof: In △ADC, we have
By Pythagoras Theorem, AC² = AD² + DC² (1)
In △ADB, we have
By Pythagoras Theorem, AB² = AD² + BD² (2)
Subtracting (2) from (1), we get
AC² - AB² = AD² + DC² - (AD² + BD²)
AC² - AB² = DC² - (DC - BC)² [∵ BD = DC - BC]
AC² - AB² = DC² - [DC² - 2DC.BC + BC²] [Using the identity (a - b)² = a² - 2ab + b²]
AC² - AB² = DC² - DC² + 2DC.BC - BC²
AC² - AB² = 2(DB + BC)BC - BC² [∵ DC = DB + BC]
AC² - AB² = 2DB.BC + 2BC² - BC²
∴ AC² = AB² + 2DB.BC + BC² [hence proved]