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Question Number 13735 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17

$$\measuredangle ABC=90,S_{ABC}=1,AB=BC$$$$\measuredangle ABD=\measuredangle DBE=\measuredangle EBC$$$$EG\mathrm{\perp }AB,DF\mathrm{\perp }BC$$$$.........................................................$$$$showthat:\frac{R_{ABC}}{R_{DHE}}=2+\sqrt{3}$$$$R_{ABC}=radiusofcircumscribedcircle$$$$oftriangleA\stackrel{\mathrm{\Delta }}{BC}andso{R}_{DHE}forD\stackrel{\mathrm{\Delta }}{HE}.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/May/17

$$whataboutinscribedcircels?$$$$isthereanyratio?$$

Commented by ajfour last updated on 23/May/17

Commented by ajfour last updated on 23/May/17

$$letAB=BC=a$$$$equationofAC:x+y=a$$$$.........ofBD:y=\frac{x}{\sqrt{3}}$$$$.........ofBE:y=\sqrt{3}x$$$$forpointD$$$$x+\frac{x}{\sqrt{3}}=a\Rightarrow x_D=\frac{a\sqrt{3}}{\sqrt{3}+1}$$$$forpointE$$$$x+x\sqrt{3}=a\Rightarrow x_E=\frac{a}{\sqrt{3}+1}$$$$x_D-x_E=\frac{a(\sqrt{3}-1)}{\sqrt{3}+1}$$$$R_{DHE}=\frac{x_D-x_E}{\sqrt{2}}=\frac{a(\sqrt{3}-1)}{\sqrt{2}(\sqrt{3}+1)}$$$$R_{ABC}=\frac{a}{\sqrt{2}}$$$$so,\frac{R_{ABC}}{R_{DHE}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}\bullet$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17

$$thankyoumr.ajfour.$$

Answered by mrW1 last updated on 22/May/17

$$letAB=BC=a$$$$\Rightarrow AC=\sqrt{2}a$$$$\frac{AD}{\sin 30}=\frac{AB}{\sin (45+30)}$$$$\Rightarrow AD=\frac{\sin 30}{\sin 75}\times a=\frac{a}{2\cos 15}=\frac{a}{2\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}}=\frac{a}{\sqrt{2+\sqrt{3}}}$$$$DE=\sqrt{2}a-\frac{2a}{\sqrt{2+\sqrt{3}}}$$$$\frac{R_{DHE}}{R_{ABC}}=\frac{DE}{AC}=1-\sqrt{\frac{2}{2+\sqrt{3}}}$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/May/17

$$1-\sqrt{\frac{2}{2+\sqrt{3}}}=1-\sqrt{\frac{2(2-\sqrt{3)}}{1}}=1-\sqrt{4-2\sqrt{3}}=$$$$1-\sqrt{{(\sqrt{3}-1)}^2}=1-(\sqrt{3}-1)=2-\sqrt{3}=\frac{1}{2+\sqrt{3}}.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17

$$thankyoumrW1.$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/May/17

$$S=\frac{1}{2}ac=2\Rightarrow ac=2\stackrel{a=c}{\Rightarrow }a=c=\sqrt{2},b=2$$$$R_{ABC}=\frac{abc}{4S}=\frac{\sqrt{2}.\sqrt{2}.2}{2\times 4}=\frac{4}{4}=1$$$$\frac{AD}{sin30}=\frac{AB}{sin(180-(30+45))}\Rightarrow \frac{AD}{\frac{1}{2}}=\frac{\sqrt{2}}{sin75}\Rightarrow$$$$AD=\frac{1}{2}\sqrt{2}\left(\frac{\sqrt{6}-\sqrt{2}}{1}\right)=\sqrt{3}-1$$$$DE=2-2AD=2-2(\sqrt{3}-1)=4-2\sqrt{3}$$$$DH=HE=\frac{DE}{\sqrt{2}}=\frac{{(\sqrt{3}-1)}^2}{\sqrt{2}}$$$$S_{DHE}=\frac{1}{2}DH.HE=\frac{1}{2}.\frac{{(\sqrt{3}-1)}^2}{\sqrt{2}}.\frac{{(\sqrt{3}-1)}^2}{\sqrt{2}}=$$$$=\frac{2(2-\sqrt{3}).2\times (2-\sqrt{3})}{4}={(2-\sqrt{3})}^2$$$$R_{DHE}=\frac{a^{{}^{\prime }}{b}^{{}^{\prime }}{c}^{{}^{\prime }}}{4S^{{}^{\prime }}}=\frac{\frac{{(\sqrt{3}-1)}^2}{\sqrt{2}}.\frac{{(\sqrt{3}-1)}^2}{\sqrt{2}}.{(\sqrt{3}-1)}^2}{4{(2-\sqrt{3})}^2}=$$$$=\frac{4{(2-\sqrt{3})}^3}{4{(2-\sqrt{3})}^2}=2-\sqrt{3}$$$$\Rightarrow \frac{R_{ABC}}{R_{DHE}}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}.◼$$

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