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Question Number 39508 by behi83417@gmail.com last updated on 06/Jul/18

Commented by behi83417@gmail.com last updated on 06/Jul/18

$$R\mathit{i}\mathit{g}\mathit{h}\mathit{t}\mathit{a}\mathit{n}\mathit{g}\mathit{e}\mathit{l}\mathit{e}:\mathit{A}{\stackrel{△}{\mathit{B}\mathit{A}}}^{\prime },(\measuredangle \mathit{B}={90}^{\bullet })\mathit{i}\mathit{s}$$$$\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{n}.$$$$\mathit{r}\mathit{e}\mathit{d}\mathit{l}\mathit{i}\mathit{n}\mathit{e}\mathit{s}\mathit{a}\mathit{r}\mathit{e}\mathit{a}\mathit{n}\mathit{g}\mathit{u}\mathit{l}\mathit{a}\mathit{r}\mathit{b}\mathit{i}\mathit{s}\mathit{e}\mathit{c}\mathit{t}\mathit{o}\mathit{r}\mathit{s}\mathit{o}\mathit{f}$$$$\measuredangle \mathit{A},\measuredangle {\mathit{A}}^{\prime }.$$$$1)\mathit{s}\mathit{h}\mathit{o}\mathit{w}\mathit{t}\mathit{h}\mathit{a}\mathit{t}:{\mathit{S}}_{\mathit{A}{\stackrel{◊}{\mathit{D}\mathit{C}\mathit{A}}}^{\prime }}=2{\mathit{S}}_{\mathit{A}{\stackrel{△}{\mathit{E}\mathit{A}}}^{\prime }}$$$$2)\frac{{\mathit{A}\mathit{A}}^{\prime }}{\mathit{D}\mathit{C}}=?,\frac{{\mathit{S}}_{{\mathit{A}\mathit{E}\mathit{A}}^{\prime }}}{{\mathit{S}}_{\mathit{D}\stackrel{△}{\mathit{E}\mathit{C}}}}=?.$$

Answered by MrW3 last updated on 07/Jul/18

Commented by MrW3 last updated on 07/Jul/18

$$\mathrm{\angle }{A}^{\prime }+\mathrm{\angle }A=90°$$$$A^{\prime }{C}^{\prime }=A^{\prime }C$$$${AD}^{\prime }=AD$$$$\Rightarrow C^{\prime }E=CE,D^{\prime }E=DE$$$$\mathrm{\angle }CED=\mathrm{\angle }{A}^{\prime }EA=180°-\frac{\mathrm{\angle }{A}^{\prime }}{2}-\frac{\mathrm{\angle }A}{2}=135°$$$$\alpha =180°-135°=45°$$$$\mathrm{\angle }{C}^{\prime }{ED}^{\prime }=135°-2\alpha =45°$$$$A_{\mathrm{\Delta }CED}=\frac{1}{2}\times CE\times DE\times \sin \mathrm{\angle }CED$$$$A_{\mathrm{\Delta }{C}^{\prime }{ED}^{\prime }}=\frac{1}{2}\times C^{\prime }E\times D^{\prime }E\times \sin \mathrm{\angle }{C}^{\prime }{ED}^{\prime }$$$$=\frac{1}{2}\times CE\times DE\times \sin 45°$$$$=\frac{1}{2}\times CE\times DE\times \sin 135°$$$$=\frac{1}{2}\times CE\times DE\times \sin \mathrm{\angle }CED$$$$=A_{\mathrm{\Delta }CED}$$$$A_{\mathrm{\Delta }{A}^{\prime }EA}=A_{\mathrm{\Delta }{A}^{\prime }{EC}^{\prime }}+A_{\mathrm{\Delta }{C}^{\prime }{ED}^{\prime }}+A_{\mathrm{\Delta }{D}^{\prime }EA}$$$$=A_{\mathrm{\Delta }{A}^{\prime }EC}+A_{\mathrm{\Delta }CED}+A_{\mathrm{\Delta }DEA}$$$$A_{A^{\prime }CDA}=A_{\mathrm{\Delta }{A}^{\prime }EA}+A_{\mathrm{\Delta }{A}^{\prime }EC}+A_{\mathrm{\Delta }CED}+A_{\mathrm{\Delta }DEA}$$$$=2A_{\mathrm{\Delta }{A}^{\prime }EA}$$

Commented by behi83417@gmail.com last updated on 07/Jul/18

$$sweetwaitingandniceproof.thanks$$$$mymaster.$$

Commented by behi83417@gmail.com last updated on 07/Jul/18

$$perpendicularfrom:Eto{AA}^{\prime },say:EH.$$$$S_{{AEA}^{\prime }}=\frac{1}{2}.EH.{AA}^{\prime }=\frac{1}{2}.r.b$$$$[\mathit{r}=\mathit{r}\mathit{a}\mathit{d}\mathit{i}\mathit{u}\mathit{s}\mathit{o}\mathit{f}\mathit{i}\mathit{n}\mathit{c}\mathit{i}\mathit{r}\mathit{c}\mathit{l}\mathit{e}\mathit{o}\mathit{f}\mathit{A}{\stackrel{△}{\mathit{B}\mathit{A}}}^{\prime }.]$$$$\frac{sin\frac{A}{2}}{A^{\prime }C}=\frac{{sinACA}^{\prime }}{b},\frac{sin\frac{A}{2}}{BC}=\frac{sinACB}{c}$$$$\frac{b}{A^{\prime }C}=\frac{c}{BC}\Rightarrow \frac{A^{\prime }C}{BC}=\frac{b}{c}\Rightarrow \frac{A^{\prime }C}{a}=\frac{b}{b+c}$$$$\Rightarrow A^{\prime }C=\frac{ab}{b+c},BC=\frac{ac}{b+c},BD=\frac{ca}{b+a},AD=\frac{cb}{b+a}$$$$S_{{ADCA}^{\prime }}=S_{{ABA}^{\prime }}-S_{DBC}=\frac{1}{2}ac-\frac{1}{2}BD.BC=$$$$=\frac{1}{2}ac-\frac{1}{2}.\frac{ac}{a+b}.\frac{ac}{b+c}=\frac{ac}{2}(1-\frac{ac}{(a+b)(b+c)})=$$$$=\frac{ac}{2}\left(\frac{ab+ac+b^2+bc-ac}{(a+b)(b+c)}\right)=\frac{ac}{2}.\frac{b^2+ab+bc}{(a+b)(b+c)}=$$$$=\frac{ac}{2}.\frac{a^2+c^2+b^2+2ba+2bc}{2(a+b)(b+c)}=\frac{ac}{2}.\frac{{(a+b+c)}^2-2ac}{2ab+2ac+2b^2+2bc}=$$$$=S.\frac{{\left(2p\right)}^2-4S}{{\left(2p\right)}^2}=S.\frac{p^2-S}{p^2}=S-\frac{S^2}{p^2}=S-r^2=$$$$=r.p-r^2=r.(p-r)=r.b=2S_{{AEA}^{\prime }}.◼$$$$note:inrightangletriangle:\mathit{b}=\mathit{p}-\mathit{r}$$$$because:p-r=p-\frac{S}{p}=\frac{p^2-S}{p}=$$$$=\frac{{(a+b+c)}^2-2ac}{4p}=\frac{a^2+b^2+c^2+2ab+2bc}{4p}$$$$=\frac{2b^2+2ba+2bc}{4p}=\frac{2b(a+b+c)}{4p}=\frac{2b.2p}{4p}=\mathit{b}.$$

Commented by behi83417@gmail.com last updated on 07/Jul/18

$${DC}^2={DB}^2+{CB}^2={\left(\frac{ac}{a+b}\right)}^2+{\left(\frac{ac}{c+b}\right)}^2=$$$$a^2{c}^2.\frac{{(b+c)}^2+{(b+a)}^2}{{(a+b)}^2{(c+b)}^2}=\frac{4S^{2}b(b+p)}{4p^4}=$$$$=\frac{S^2}{p^2}.b.\frac{b+p}{p^2}=b.r^2.\frac{2p-r}{p^2}=b.\frac{r^2}{p^2}(2p-r)$$$$\Rightarrow \frac{DC}{{AA}^{\prime }}=\frac{r^2}{p^2}(2p-r).$$$$note:$$$${(b+c)}^2+{(b+a)}^2=3b^2+2bc+2ab=$$$$b(3b+a+c)=b(2b+2p)=2b(b+p)$$$$(a+b)(c+b)=ac+ab+bc+b^2=$$$$=\frac{a^2+b^2+c^2+2ac+2bc+2ab}{2}=\frac{{\left(2p\right)}^2}{2}=2p^2$$

Commented by behi83417@gmail.com last updated on 08/Jul/18

$$S_{ACE}=\frac{1}{2}.r.A^{\prime }C=\frac{1}{2}.r.\frac{ab}{b+c}$$$$S_{ADE}=\frac{1}{2}.r.AD=\frac{1}{2}.r.\frac{c.b}{a+b}$$$$S_{DEC}=2S_{{AEA}^{\prime }}-S_{{AEA}^{\prime }}-\frac{b.r}{2}(\frac{a}{b+c}+\frac{c}{a+b})=$$$$=\frac{b.r}{2}(1-\frac{a}{b+c}-\frac{c}{a+b})=\frac{br(ab+ac+bc+b^2-a^2-ab-c^2-cb)}{2(b+c)(b+a)}=$$$$=\frac{br.ac}{4p^2}=\frac{br.2S}{4p^2}=\frac{b}{2p}$$$$\Rightarrow \frac{S_{{AEA}^{\prime }}}{S_{DRC}}=\frac{\frac{br}{2}}{\frac{b}{2p}}=r.p(????????????\to \mathrm{\infty })$$

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