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Question Number 193293 by TUN last updated on 09/Jun/23

Answered by deleteduser1 last updated on 09/Jun/23

$$h=a+b+c$$$$\mid AH{\mid }^2=\mid b+c{\mid }^2=\frac{{(b+c)}^2}{bc}=(b+c)(\stackrel{-}{b}+\stackrel{-}{c})$$$$\mid BC{\mid }^2=\mid c-b{\mid }^2=(c-b)(\stackrel{-}{c}-\stackrel{-}{b})=(c-b)\left(\frac{b-c}{bc}\right)=\frac{-{(b-c)}^2}{bc}$$$$\frac{\mid AH{\mid }^2}{\mid BC{\mid }^2}=\frac{-{(b+c)}^2}{{(b-c)}^2}=3(wherec={be}^{i\frac{\pi }{3}})$$$$x^2{R}^2=9R^2\Rightarrow x^2=9\Rightarrow x=3$$

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