Question

In triangle ABC; AD⊥BC. Prove that AC²= AB² + BC²-2BCBD.



FIGURE IS GIVEN IN ATTACHMENT

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

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Answer 2

Step-by-step explanation:

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