Question

In a triangle ABC AD is perpendicular to BC. Prove that AC^2=AB^2+BC^2-2BC*BD

Answer 1

In the right angle triangle ABD, we have
         AD² = AB² - BD²
In the right angle triangle  ACD we have
        AD² = AC² - CD²

So    AC² - CD²  = AB² - BD²
        =>  AC² = AB² + CD²  -  BD²
        =>         = AB² + (CD + BD)* (CD - BD )
                     = AB² + BC * [(BC - BD) - BD]
                     = AB² + BC * [ BC - 2 BD ]
                       = AB² +BC² - 2 BC * BD

Answer 2

hope this would helps you