Question

29. In figure, angleB < 90° and AD I BC. Prove that:
AC2 = AB + BC-2BC.BD​

Answer 1

In the right angle triangle ABD, we have AD2 = AB? - BD² In the right angle triangle ACD we have AD2 = AC? CD²

So AC? - CD? = AB? - BD? => AC? = AB? + CD? - BD?

= AB? + (CD + BD)* (CD-BD )

= AB? + BC * [(BC BD)- BD] = AB? + BC * [ BC - 2 BD ]

= AB? +BC? - 2 BC * BD