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Question Number 12725 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Apr/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Apr/17

$$intriangle:ABC,pointDlocatedon$$$$triangleplan,suchthat:$$$$\mathrm{\angle }ADB=\mathrm{\angle }BDC=\mathrm{\angle }CDAand:$$$$DH\parallel AB,DG\parallel BC,DF\parallel AC.$$$$1)find:\frac{DH}{AB}+\frac{DG}{BC}+\frac{DF}{AC}.$$$$2)prove:DH+DG+DF<DA+DB+DC$$$$3)\frac{S_{ADG}}{S_{DBH}}=?$$

Commented by mrW1 last updated on 01/May/17

Commented by mrW1 last updated on 01/May/17

$$whatdoyoumeaninquestion3?$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/May/17

$$hellodearmrW1.thankyousomuch$$$$forsolvingthisquestion.youare$$$$numberoneintinkutara.$$$$aboutyourcommunet:$$$$S_{ADG}=areaoftriangleADGandso.$$

Answered by mrW1 last updated on 01/May/17

$$\mathit{Q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}1)$$$$$$$$since\measuredangle ADB=\measuredangle ADC=\measuredangle BDC$$$$\Rightarrow \measuredangle ADB=\measuredangle ADC=\measuredangle BDC=\frac{360}{3}=120°$$$$\sin 120°=\frac{\sqrt{3}}{2}$$$$\cos 120°=-\frac{1}{2}$$$$$$$$let\overline{DA}=a$$$$let\overline{DB}=b$$$$let\overline{DC}=c$$$$$$$${AC}^2=a^2+c^2-2ac\times \cos \measuredangle ADC=a^2+c^2+ac$$$$\Rightarrow AC=\sqrt{a^2+c^2+ac}$$$$similarily$$$$\Rightarrow AB=\sqrt{a^2+b^2+ab}$$$$\Rightarrow BC=\sqrt{b^2+c^2+bc}$$$$$$$$in\mathrm{\Delta }BDC:$$$$\frac{BC}{\sin 120°}=\frac{b}{\sin \beta }=\frac{c}{\sin (60-\beta )}...\left(i\right)$$$$$$$$in\mathrm{\Delta }ADC:$$$$\frac{AC}{\sin 120°}=\frac{a}{\sin \alpha }=\frac{c}{\sin (60-\alpha )}...\left(ii\right)$$$$$$$$from\left(i\right):$$$$\frac{c}{b}\sin \beta =\sin (60-\beta )=\sin 60\cos \beta -\cos 60\sin \beta$$$$\frac{c}{b}\sin \beta =\frac{\sqrt{3}}{2}\cos \beta -\frac{1}{2}\sin \beta$$$$\Rightarrow \cos \beta =\frac{2c+b}{\sqrt{3}b}\sin \beta$$$$similarilyfrom\left(ii\right):$$$$\Rightarrow \cos \alpha =\frac{2c+a}{\sqrt{3}a}\sin \alpha$$$$$$$$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\sin \beta \cos \alpha$$$$=\sin \alpha \times \frac{2c+b}{\sqrt{3}b}\sin \beta +\sin \beta \times \frac{2c+a}{\sqrt{3}a}\sin \alpha$$$$=\frac{2}{\sqrt{3}}\times \frac{ab+bc+ca}{ab}\times \sin \alpha \sin \beta$$$$$$$$in\mathrm{\Delta }CDG:$$$$\frac{DG}{\sin \alpha }=\frac{DC}{\sin \mathrm{\angle }CGD}=\frac{c}{\sin (180-\alpha -\beta )}=\frac{c}{\sin (\alpha +\beta )}$$$$\Rightarrow DG=c\times \frac{\sin \alpha }{\sin (\alpha +\beta )}=\frac{c\times \sin \alpha }{\frac{2}{\sqrt{3}}\times \frac{ab+bc+ca}{ab}\times \sin \alpha \sin \beta }$$$$=\frac{\sqrt{3}}{2}\times \frac{b}{\sin \beta }\times \frac{ac}{ab+bc+ca}$$$$$$$$from\left(i\right)wehave$$$$\frac{b}{\sin \beta }=\frac{BC}{\sin 120°}=\frac{BC}{\frac{\sqrt{3}}{2}}$$$$\Rightarrow \frac{\sqrt{3}}{2}\times \frac{b}{\sin \beta }=BC$$$$$$$$\Rightarrow DG=BC\times \frac{ac}{ab+bc+ca}$$$$or$$$$\frac{DG}{BC}=\frac{ac}{ab+bc+ca}$$$$$$$$similarily$$$$\frac{DH}{AB}=\frac{bc}{ab+bc+ca}$$$$\frac{DF}{AC}=\frac{ab}{ab+bc+ca}$$$$$$$$\frac{DH}{AB}+\frac{DF}{AC}+\frac{DG}{BC}=\frac{bc+ab+ac}{ab+bc+ca}=1$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/May/17

$$sobeautiful.thanksalot.$$

Answered by mrW1 last updated on 01/May/17

$$\mathit{Q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}2)$$$$$$$$seeabove:$$$$DG=\frac{ac}{ab+bc+ca}\times BC$$$$=\frac{ac}{ab+bc+ca}\times \sqrt{b^2+c^2+bc}$$$$<\frac{ac}{ab+bc+ca}\times \sqrt{b^2+c^2+2bc}$$$$=\frac{ac(b+c)}{ab+bc+ca}$$$$\Rightarrow DG<\frac{ac(b+c)}{ab+bc+ca}$$$$$$$$similarily$$$$DH<\frac{bc(a+b)}{ab+bc+ca}$$$$DF<\frac{ab(c+a)}{ab+bc+ca}$$$$$$$$DG+DH+DF<\frac{ac(b+c)+bc(a+b)+ab(c+a)}{ab+bc+ca}$$$$=\frac{ac(a+b+c)+bc(a+b+c)+ab(c+a+b)-a^{2}c-c^{2}b-b^{2}a}{ab+bc+ca}$$$$=\frac{(ac+bc+ab)(a+b+c)-a^{2}c-c^{2}b-b^{2}a}{ab+bc+ca}$$$$=a+b+c-\frac{a^{2}c+c^{2}b+b^{2}a}{ab+bc+ca}$$$$<a+b+c$$$$=DA+DB+DC$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/May/17

$$perfectandnice.thankyouverymuch.$$

Answered by mrW1 last updated on 01/May/17

$$\mathit{Q}\mathit{u}\mathit{e}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}3)$$$$$$$$\frac{GC}{AC}=\frac{DH}{AB}=\frac{bc}{ab+bc+ca}$$$$$$$$S_{\mathrm{\Delta }ADC}=\frac{1}{2}\times ac\times \sin 120°=\frac{\sqrt{3}}{4}ac$$$$S_{\mathrm{\Delta }GDC}=\frac{GC}{AC}\times S_{\mathrm{\Delta }ADC}=\frac{bc}{ab+bc+ca}\times \frac{\sqrt{3}ac}{4}$$$$S_{\mathrm{\Delta }ADG}=(1-\frac{GC}{AC})\times S_{\mathrm{\Delta }ADC}=(1-\frac{bc}{ab+bc+ca})\times \frac{\sqrt{3}ac}{4}=\frac{\sqrt{3}{a}^{2}c(b+c)}{4(ab+bc+ca)}$$$$$$$$similarily$$$$S_{\mathrm{\Delta }BDH}=\frac{DF}{AC}\times S_{\mathrm{\Delta }BDC}=\frac{ab}{ab+bc+ca}\times \frac{\sqrt{3}bc}{4}=\frac{\sqrt{3}{ab}^{2}c}{4(ab+bc+ca)}$$$$$$$$\Rightarrow \frac{S_{\mathrm{\Delta }ADG}}{S_{\mathrm{\Delta }BDH}}=\frac{a^{2}c(b+c)}{{ab}^{2}c}=\frac{a(b+c)}{b^2}=\frac{a}{b}(1+\frac{c}{b})$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/May/17

$$S_{BDG}=\frac{\sqrt{3}}{4}{b}^2.\frac{ac}{ab+bc+ac}$$$$S_{ADF}=\frac{\sqrt{3}}{4}{a}^2.\frac{bc}{ab+bc+ac}$$$$S_{CDG}=\frac{\sqrt{3}}{4}{c}^2.\frac{ab}{ab+bc+ac}$$$$\frac{S_{ADF}}{a^2}+\frac{S_{BDG}}{b^2}+\frac{S_{CDG}}{c^2}=\frac{\sqrt{3}}{4}.$$$$.........................................................$$$$\frac{S_{ADG}}{ac}=\frac{\sqrt{3}}{4}.\frac{ba+ca}{ab+bc+ac}$$$$\frac{S_{BDF}}{ba}=\frac{\sqrt{3}}{4}.\frac{bc+ab}{ab+bc+ac}$$$$\frac{S_{CDG}}{bc}=\frac{\sqrt{3}}{4}.\frac{ac+bc}{ab+bc+ac}$$$$\frac{S_{ADG}}{ac}+\frac{S_{BDF}}{ab}+\frac{S_{CDG}}{bc}=\frac{\sqrt{3}}{2}.$$$$........................................................$$$$S_{ABC}=\frac{\sqrt{3}}{4}(ab+bc+ac).$$

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