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In Δ ABC (Figure 3), AD丄BC. Prove thatAC2 = AB2 + BC2-2BOx BDB D |
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Answer» In the right angle triangle ABD, we have AD² = AB² - BD²In 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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