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由于$BEFGD$五点共圆。所以$\frac{\pi}{2}=\angle{EBD}=\angle{FED}=\angle{EGD}$
注意到
\[\frac{GD}{GE}=\frac{AB}{BC}=\frac{3}{4}\]\[\iff\Delta EGD\sim\Delta ABC\iff\angle{GED}=\angle{DFG}=\angle{C}\\\iff DF=DC,\angle{EDG}=\frac{\pi}{2}-\angle{EFG}=\angle{A}\iff AE=EF\]
设$BE=y,BD=x, AE=3-y, CD=4-x$. 联立方程
\[\left\{ \begin{array}{rcl}\frac{\sqrt{x^2+y^2}}{2}=4-x
& \\
\frac{\sqrt{x^2+y^2}}{2}=3-y\end{array}\right.\]
解得
\[\left\{ \begin{array}{rcl} x=\frac{31}{12}
& \\
y=\frac{17}{14}\end{array}\right.\]
于是\[DE=\sqrt{x^2+y^2}=\sqrt{\left(\frac{31}{14}\right)^2+\left(\frac{17}{14}\right)^2}=\frac{25\sqrt{2}}{14}\] |
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其妙
Posted at 2014-3-15 12:50:47
kuing
Posted at 2014-3-19 00:50:05
