Question

In triangle ABC, AD perpendicular on BC show that AC ²=AB²+BC²-2BC*BD

Answer 1

Step-by-step explanation:

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 x [(BC - BD) - BD]

           = AB² + BC x [ BC - 2 BD ]

            = AB² +BC² - 2 BC x BD

Hence Proved!

Hope this helps!

If it does then please mark it as Brainliest!

Regards,

Soham Patil.