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Question Number 200471 by Mingma last updated on 19/Nov/23

Answered by deleteduser1 last updated on 19/Nov/23

$$\frac{{CB}_1}{B_{1}A}\times \frac{1}{2}\times \frac{1}{4}=1\Rightarrow \frac{{CB}_1}{B_{1}A}=8$$$$\frac{\left[CAB\right]}{\left[ADB\right]}=\frac{{CC}_1}{{DC}_1}=\frac{CD}{{DC}_1}+1=\frac{{CB}_1}{B_{1}A}+\frac{{CA}_1}{A_{1}B}+1=13$$$$\Rightarrow \left[ADB\right]=\frac{65}{13}=5$$$$\frac{\left[CDA\right]}{\left[CDB\right]}=\frac{1}{2}\Rightarrow \left[CDA\right]+\left[CDB\right]=3\left[CDA\right]=65-5$$$$\Rightarrow \left[CDA\right]=20;\frac{\left[B_{1}CD\right]}{\left[B_{1}AD\right]}=8$$$$\left[B_{1}CD\right]+\left[B_{1}AD\right]=\frac{9\left[B_{1}CD\right]}{8}=20\Rightarrow \left[B_{1}CD\right]=\frac{160}{9}$$

Commented by Rupesh123 last updated on 19/Nov/23

Nice solution,sir!

Commented by klipto last updated on 28/Nov/23

160/9

Answered by mr W last updated on 19/Nov/23

$$\overrightarrow{AB}=\mathit{a}$$$$\overrightarrow{BC}=\mathit{b}$$$$\overrightarrow{CA}=-\mathit{a}-\mathit{b}$$$$\overrightarrow{{AA}_1}=\mathit{a}+\frac{1}{5}\mathit{b}$$$$\overrightarrow{{CC}_1}=-\frac{2}{3}\mathit{a}-\mathit{b}$$$$sayAD={sAA}_1=s(\mathit{a}+\frac{1}{5}\mathit{b})$$$$CD={tCC}_1=t(-\frac{2}{3}\mathit{a}-\mathit{b})$$$$AD=\mathit{a}+\mathit{b}+CD$$$$s(\mathit{a}+\frac{1}{5}\mathit{b})=\mathit{a}+\mathit{b}+t(-\frac{2}{3}\mathit{a}-\mathit{b})$$$$(1-s-\frac{2t}{3})\mathit{a}+(1-\frac{s}{5}-t)\mathit{b}=0$$$$1-s-\frac{2t}{3}=0...\left(i\right)$$$$1-\frac{s}{5}-t=0...\left(ii\right)$$$$\Rightarrow s=\frac{5}{13},t=\frac{12}{13}$$$$BD=-\mathit{a}+AD=-\mathit{a}+\frac{5}{13}(\mathit{a}+\frac{1}{5}\mathit{b})=-\frac{8}{13}\mathit{a}+\frac{1}{13}\mathit{b}$$$$sayBD={uBB}_1,{CB}_1=vCA$$$$v(-\mathit{a}-\mathit{b})=-\mathit{b}+\frac{1}{u}(-\frac{8}{13}\mathit{a}+\frac{1}{13}\mathit{b})$$$$(\frac{8}{13u}-v)\mathit{a}+(1-\frac{1}{13u}-v)\mathit{b}=0$$$$\frac{8}{13u}-v=0...\left(iii\right)$$$$1-\frac{1}{13u}-v=0...\left(iv\right)$$$$\Rightarrow u=\frac{9}{13},v=\frac{8}{9}$$$$$$$${\mathrm{\Delta }}_{B_{1}CD}=(1-u){\mathrm{\Delta }}_{B_{1}CB}=(1-u)v{\mathrm{\Delta }}_{ABC}$$$${\mathrm{\Delta }}_{B_{1}CD}=(1-\frac{9}{13})\times \frac{8}{9}{\mathrm{\Delta }}_{ABC}=\frac{32}{117}{\mathrm{\Delta }}_{ABC}$$$$=\frac{32\times 65}{117}=\frac{160}{9}\approx 17.8$$

Commented by Rupesh123 last updated on 19/Nov/23

Nice solution,sir!

Commented by Anonim_X last updated on 19/Nov/23

$$greatsolution$$

Commented by mr W last updated on 20/Nov/23

$$certainlywecandirectlyapplythe$$$${Ceva}^{\prime }stheorem,seebelow.buthere$$$$iappliedthevectormethod.$$

Commented by mr W last updated on 20/Nov/23

Answered by ajfour last updated on 19/Nov/23

$${△}_0=\frac{5\times 3}{2}\sin B=6.5\Rightarrow \sin B=\frac{13}{15}$$$$$$$$letD(0,0);A(-a,0);C(0,b)$$$$y_{B1}=-\frac{b}{4}$$$$\sin B=\frac{(b+\frac{b}{4})}{5}=\frac{13}{15}$$$$\Rightarrow b=\frac{52}{15}$$$$\cos B=\left(\frac{9+25-{AC}^2}{2\times 3\times 5}\right)=\sqrt{1-{\left(\frac{13}{15}\right)}^2}$$$${AC}^2=34-2\sqrt{56}=a^2+b^2$$$$x_{C1}=0=\frac{-2a+x_B}{3}\Rightarrow x_B=2a$$$$eqofACy=\frac{b}{a}x+b$$$$eqof{BB}_{1}y=\left(\frac{y_B}{x_B}\right)x=-\frac{b}{8a}x$$$$intersection(x_{B1},y_{B1})$$$$\frac{b}{a}x+b=-\frac{b}{8a}x$$$$\Rightarrow x_{B1}=-\frac{8a}{9}\Rightarrow y_{B1}=\frac{b}{9}$$$${△}_{B_{1}CD}=\frac{b(-x_{B_1})}{2}=\frac{4ab}{9}$$$$=\frac{4b\sqrt{(a^2+b^2)-b^2}}{9}$$$$=\frac{4}{9}\left(\frac{52}{15}\right)\sqrt{34-2\sqrt{56}-{\left(\frac{52}{15}\right)}^2}$$$$\approx 4.0809$$

Commented by mr W last updated on 19/Nov/23

$$ithinkinthediagram1,2and1,4$$$$justmeantheratiosare1:2and1:4.$$

Commented by ajfour last updated on 19/Nov/23

$$oh!thankssir.dintdawnonmeso..$$

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