回復 4# beaglewu 的帖子
第 4 題
\(\begin{align}
& \overline{AF}=x,\overline{BF}=1 \\
& \Delta AFG=\Delta FBD=a \\
& \frac{\Delta AFG}{\Delta DFG}=\frac{\overline{AG}}{\overline{DG}}=\frac{\overline{AF}}{\overline{BF}}=x \\
& \Delta DFG=\frac{a}{x} \\
& \frac{\Delta AFD}{\Delta BFD}=\frac{\overline{AF}}{\overline{BF}} \\
& \frac{a+\frac{a}{x}}{a}=\frac{x}{1} \\
& 1+\frac{1}{x}=x \\
& \frac{\overline{AF}}{\overline{BF}}\text{=}x=\frac{1+\sqrt{5}}{2} \\
\end{align}\)
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2019-1-31 22:49