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Question Number 201322 by cherokeesay last updated on 04/Dec/23

Answered by mr W last updated on 05/Dec/23

$$BD=x$$$${BA}^2=x^2-1^2$$$$BC=\sqrt{2^2+x^2-1^2}=\sqrt{x^2+3}$$$$\frac{BE}{BD}=\frac{EC}{DC}\Rightarrow \frac{BE}{x}=\frac{EC}{1}\Rightarrow BE=xEC$$$$BE+EC=(x+1)EC=\sqrt{x^2+3}$$$$\Rightarrow EC=\frac{\sqrt{x^2+3}}{x+1}$$$$EC=2\times 1\cos C=2\times \frac{1+1}{\sqrt{x^2+3}}=\frac{4}{\sqrt{x^2+3}}$$$$\frac{4}{\sqrt{x^2+3}}=\frac{\sqrt{x^2+3}}{x+1}$$$$\Rightarrow x^2+3=4(x+1)$$$$\Rightarrow x^2-4x-1=0$$$$\Rightarrow x=2+\sqrt{5}✓$$

Commented by cherokeesay last updated on 05/Dec/23

$$thankyousir!$$

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