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Question Number 206640 by cortano21 last updated on 21/Apr/24

Answered by mr W last updated on 21/Apr/24

Commented by mr W last updated on 21/Apr/24

$$\frac{EY}{YB}\times \frac{x}{2}\times \frac{y+z}{z}=1$$$$\Rightarrow \frac{y}{z}=\frac{2}{x}-1=\frac{2-x}{x}$$$$\frac{FZ}{ZC}\times \frac{y}{z}\times \frac{p+q}{q}=1$$$$\frac{2-x}{x}\times (1+\frac{p}{q})=1$$$$\Rightarrow \frac{p}{q}=\frac{x}{2-x}-1=\frac{2(x-1)}{2-x}$$$$\frac{DX}{XA}\times \frac{p}{q}\times \frac{x+2}{2}=1$$$$\frac{2(x-1)}{2-x}\times \frac{x+2}{2}=1$$$$x^2+2x-4=0$$$$\Rightarrow x=\sqrt{5}-1✓$$

Answered by A5T last updated on 21/Apr/24

$$\frac{BD}{DC}=\frac{k}{2};\frac{BC}{DC}=\frac{k+2}{2}$$$$\frac{AF}{FB}\times \frac{BC}{CD}\times \frac{DX}{AX}=1\Rightarrow \frac{AF}{FB}=\frac{2}{k+2}\Rightarrow \frac{FB}{BA}=\frac{k+2}{4+k}$$$$\frac{AE}{EC}\times \frac{CZ}{ZF}\times \frac{FB}{BA}=1\Rightarrow \frac{AE}{EC}=\frac{4+k}{k+2}$$$$\frac{BD}{DC}\times \frac{CA}{AE}\times \frac{EY}{EB}=1\Rightarrow \frac{\left(k\right)(6+2k)}{2(4+k)}=1$$$$\Rightarrow 6k+2k^2=8+2k\Rightarrow k^2-2k-4=0\Rightarrow k=\sqrt{5}-1$$

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