Question

In triangle ABc seg BD perpendicular side AC
C-D-A. Prove that : AB ² =
that : AB² = BC² + AC² – BC.AC

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Answer 1

Answer:

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

Step-by-step explanation: