Question

In triangle ABC AD perpendicular to BC . Prove that AC2 = AB2 + BC2-2 BC.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 * [(BC - BD) - BD]

= AB² + BC * [ BC - 2 BD ]

= AB² +BC² - 2 BC * BD

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