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Question Number 104998 by mathocean1 last updated on 25/Jul/20

Commented by mathocean1 last updated on 25/Jul/20

$$ABCDandDEFGaresquares.$$$$AB=2DE.$$$$Determinate\alpha .$$

Commented by ajfour last updated on 25/Jul/20

$$manyunnecessaryconstructions$$$$makequestionconfusing..$$

Answered by 1549442205PVT last updated on 25/Jul/20

Commented by 1549442205PVT last updated on 25/Jul/20

$$$$$$\mathrm{Putting}\mathrm{AB}=2\mathrm{a}\Rightarrow \mathrm{DE}=\mathrm{a}.\mathrm{Denote}\mathrm{by}\mathrm{I}$$$$\mathrm{intersection}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{ray}\mathrm{AE}\mathrm{and}$$$$\mathrm{the}\mathrm{circle}\left(\mathrm{O}\right)$$$$\mathrm{From}{\mathrm{Pithago}}^{\prime }\mathrm{s}\mathrm{theorem}\mathrm{we}\mathrm{get}$$$$\mathrm{AE}=\sqrt{{\mathrm{AD}}^2+{\mathrm{DE}}^2}=\mathrm{a}\sqrt{5}.\left(1\right)$$$$\mathrm{From}\mathrm{the}\mathrm{relation}\mathrm{formular}\mathrm{in}\mathrm{circle}\mathrm{we}$$$$\mathrm{get}\mathrm{EC}.\mathrm{ED}=\mathrm{EA}.\mathrm{EI}\iff {\mathrm{a}}^2=\mathrm{a}\sqrt{5}.\mathrm{EI}$$$$\Rightarrow \mathrm{EI}=\frac{\mathrm{a}}{\sqrt{5}}.$$$$\mathrm{we}\mathrm{have}\frac{\mathrm{EI}}{\mathrm{ED}}=\frac{\mathrm{EF}}{\mathrm{AE}}=\frac{1}{\sqrt{5}}.\mathrm{It}\mathrm{follows}\mathrm{that}$$$$\mathrm{\Delta }\mathrm{AOE}\backsimeq \mathrm{\Delta }\mathrm{FIE}(\mathrm{s}.\mathrm{a}.\mathrm{s})\Rightarrow \widehat{\mathrm{EIF}}=\widehat{\mathrm{AOE}}=135°$$$$\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hands},\widehat{\mathrm{AIB}}=\widehat{\mathrm{ADB}}=45°$$$$\Rightarrow \widehat{\mathrm{AIB}}+\widehat{\mathrm{EIF}}=45°+135°=180°.\mathrm{This}$$$$\mathrm{shows}\mathrm{that}\mathrm{E},\mathrm{I},\mathrm{B}\mathrm{are}\mathrm{colinear}(\mathrm{q}.\mathrm{e}.\mathrm{d})$$

Commented by mathocean1 last updated on 25/Jul/20

$$thankyousir.$$

Answered by mr W last updated on 25/Jul/20

Commented by mr W last updated on 25/Jul/20

$$\tan \beta =\frac{1}{3}$$$$\tan \gamma =\frac{1}{2}$$$$\alpha =\beta +\gamma$$$$\tan \alpha =\tan (\beta +\gamma )=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\times \frac{1}{2}}=1$$$$\Rightarrow \alpha =45°$$$$\mathrm{\angle }AIB=\alpha =\mathrm{\angle }ACB=45°$$$$\Rightarrow Iisonthecircle.$$$$$$$$BDisdiameterofcircle,$$$$\Rightarrow \mathrm{\Delta }BDIisrightangledtriangle,$$$$\Rightarrow DI\mathrm{\perp }BI$$

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