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Question Number 163610 by amin96 last updated on 08/Jan/22

Answered by mr W last updated on 08/Jan/22

Commented by mr W last updated on 08/Jan/22

$$\beta +\frac{\alpha }{2}=90°$$$$\beta +\gamma =90°$$$$\Rightarrow \gamma =\frac{\alpha }{2}$$$$BC=2(a+b)\sin \frac{\alpha }{2}$$$$BD=2(a+b)\sin \frac{\alpha }{2}+a$$$$BD\times \sin \gamma =(2(a+b)\sin \frac{\alpha }{2}+a)\sin \gamma =b$$$$(1-\cos \alpha +\sin \frac{\alpha }{2})a=b\cos \alpha$$$$\frac{b}{a}=\frac{1-\cos \alpha +\sin \frac{\alpha }{2}}{\cos \alpha }$$$$ED=\frac{b}{\tan \gamma }$$$$FD=\frac{b}{\tan \gamma }-a\tan \alpha$$$$\frac{FD}{a}=\frac{1-\cos \alpha +\sin \frac{\alpha }{2}}{\cos \alpha \tan \gamma }-\tan \alpha$$$$$$$$\left[AEF\right]=\frac{a\times a\tan \alpha }{2}$$$$\left[FCD\right]=\frac{a\times FD\sin \gamma }{2}$$$$\lambda =\frac{\left[AEF\right]}{\left[FCD\right]}=\frac{a\tan \alpha }{FD\sin \gamma }$$$$\lambda =\frac{\tan \alpha }{(\frac{1-\cos \alpha +\sin \frac{\alpha }{2}}{\cos \alpha \tan \gamma }-\tan \alpha )\sin \gamma }$$$$\lambda =\frac{\tan 2\gamma }{(\frac{1-\cos 2\gamma +\sin \gamma }{\cos 2\gamma \tan \gamma }-\tan 2\gamma )\sin \gamma }$$$$\lambda =\frac{1}{(\frac{1-\cos 2\gamma +\sin \gamma }{\sin 2\gamma \tan \gamma }-1)\sin \gamma }$$$$\lambda =\frac{\sin 2\gamma }{2{\sin }^2\gamma \cos \gamma +\sin \gamma \cos \gamma -\sin 2\gamma \sin \gamma }$$$$\lambda =2$$$$\Rightarrow \frac{\left[AEF\right]}{\left[FCD\right]}=2,independentlyfromthe$$$$shapeofthetriangleABC!$$

Commented by mr W last updated on 08/Jan/22

Commented by Tawa11 last updated on 08/Jan/22

$$\mathrm{Wow},\mathrm{great}\mathrm{sir}.$$

Commented by amin96 last updated on 08/Jan/22

$$\mathbf{bravo}\mathbf{master}\mathbf{bravooo}$$

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