ABCD为正方形，E,F在CD,BC上，∠EAF=45°，AE,AF交BD于E1,F1，求证：SΔAEF=2SΔAE1F1

luyuanhong

下面是我的另一种证法，不用添加任何辅助线。

kanyikan

王守恩

luyuanhong 发表于 2021-12-24 10:28
下面是我的另一种证法，不用添加任何辅助线。

\(记∠BAF=\pi/8+\theta\ \ \ \ \ ∠DAE=\pi/8-\theta\)

\(ΔAEF相似ΔAE_{1}F_{1},由下得\frac{EF}{F_{1}E_{1}}=\frac{\sin(\pi/4)}{\sin^2(\pi/4)}=\frac{\sqrt{2}}{1}\)

\(1,在ΔEFA中,\ \ \ EF=\sin(\pi/4)\ \ \ \ \ \ FA =\cos(\pi/8-\theta)\)

\(2,在ΔFAB中,\ \ \ FA=\cos(\pi/8-\theta)\ \ \ \ \ AB=\cos(\pi/8-\theta)\cos(\pi/8+\theta)\)

\(3,在ΔBAF_{1}中,\ \ BA=\cos(\pi/8-\theta)\cos(\pi/8+\theta)\ \ \ \ AF_{1}=\cos(\pi/8+\theta)\sin(\pi/4)\)

\(4,在ΔAF_{1}E_{1}中,\ \ AF_{1}=\cos(\pi/8+\theta)\sin(\pi/4)\ \ \ \ \ \ F_{1}E_{1}=\sin^2(\pi/4)\)

王守恩

luyuanhong 发表于 2021-12-23 17:30

我们有的是方法!   3楼的图。谢谢陆老师！

\(ΔAEF相似ΔAE_{1}F_{1},由下得\frac{AF}{E_{1}A}=\frac{\frac{\sin(\pi/4+\theta)}{\sin(\pi/4)}}{\sin(\pi/4+\theta)}=\frac{\sqrt{2}}{1}\)

\(1,在ΔE_{1}AE_{1}中,\ \ \ E_{1}A =\sin(\pi/4+\theta)\ \ \ AF_{1}=\cos(\theta)\)

\(2,在ΔF_{1}AB中,\ \ \ F_{1}A=\cos(\theta)\ \ \ AB=\frac{\sin(\pi/4+\theta)cos(\theta)}{\sin(\pi/4)}\)

\(3,在ΔBAF中,\ \ \ BA=\frac{\sin(\pi/4+\theta)cos(\theta)}{\sin(\pi/4)}\ \ \ AF=\frac{\sin(\pi/4+\theta)}{\sin(\pi/4)}\)


\(ΔAEF相似ΔAE_{1}F_{1},由下得\frac{AE}{F_{1}A}=\frac{\frac{cos(\theta)}{\sin(\pi/4)}}{cos(\theta)}=\frac{\sqrt{2}}{1}\)

\(1,在ΔF_{1}AE_{1}中,\ \ \ F_{1}A=\cos(\theta)\ \ \ AE_{1}=\sin(\pi/4+\theta)\)

\(2,在ΔE_{1}AD中,\ \ \ E_{1}A=\sin(\pi/4+\theta)\ \ \ AD=\frac{\sin(\pi/4+\theta)cos(\theta)}{\sin(\pi/4)}\)

\(3,在ΔDAE中,\ \ \ DA=\frac{\sin(\pi/4+\theta)cos(\theta)}{\sin(\pi/4)}\ \ \ AE=\frac{cos(\theta)}{\sin(\pi/4)}\)



\(ΔAEF相似ΔAE_{1}F_{1},由下得\frac{AF}{E_{1}A}=\frac{1}{\sin(\pi/4)}=\frac{\sqrt{2}}{1}\)

\(1,在ΔE_{1}AD中,\ \ \ E_{1}A=\sin(\pi/4)\ \ \ AD=\cos(\theta)=AB\)

\(2,在ΔBAF中,\ \ \ BA=\cos(\theta)\ \  \ AF=1\)

........

波斯猫猫

可以证明：只有当∠EAF=45°时，对角线BD才平分动ΔEAF。

王守恩

luyuanhong 发表于 2021-12-24 10:28
下面是我的另一种证法，不用添加任何辅助线。

来只纸老虎。谢谢陆老师！12楼的图。\(\frac{S_{ΔABF_{1}}+S_{ΔADE_{1}}}{S_{ΔFBF_{1}}+S_{ΔEDE_{1}}+S_{ΔCFE}}\)

\(=\frac{1*\frac{\sin(\pi/8+\theta)}{\cos(\pi/8-\theta)}*\sin(\pi/4)+1*\frac{\sin(\pi/8-\theta)}{\cos(\pi/8+\theta)}*\sin(\pi/4)}{\frac{\sin(\pi/8+\theta)}{\cos(\pi/8+\theta)}*\frac{\sin(\pi/8+\theta)}{\cos(\pi/8-\theta)}*\sin(\pi/4)+\frac{\sin(\pi/8-\theta)}{\cos(\pi/8-\theta)}*\frac{\sin(\pi/8-\theta)}{\cos(\pi/8+\theta)}*\sin(\pi/4)+(1-\frac{\sin(\pi/8+\theta)}{\cos(\pi/8+\theta)})*(1-\frac{\sin(\pi/8-\theta)}{\cos(\pi/8-\theta)})*\sin(\pi/2)}=1\)

王守恩

王守恩 发表于 2021-12-25 13:59
来只纸老虎。谢谢陆老师！12楼的图。\(\frac{S_{ΔABF_{1}}+S_{ΔADE_{1}}}{S_{ΔFBF_{1}}+S_{ΔEDE_{1}} ...

谢谢陆老师！12楼的图。

\(记∠BAF=\pi/8+\theta\ \ \ ∠DAE=\pi/8-\theta\ \ \ \ AB=1\)

\(\frac{2S_{ΔAEF}}{2SΔAE_{1}F_{1}}=\frac{AE*AF*\sin(\pi/4)}{AE_{1}*AF_{1}*\sin(\pi/4)}=\frac{\frac{1}{\cos(\pi/8-\theta)}*\frac{1}{\cos(\pi/8+\theta)}*\sin(\pi/4)}{\frac{\sin(\pi/4)}{\cos(\pi/8+\theta)}*\frac{\sin(\pi/4)}{\cos(\pi/8-\theta)}*\sin(\pi/4)}=2\)