Menu Close

Question-206640




Question Number 206640 by cortano21 last updated on 21/Apr/24
Answered by mr W last updated on 21/Apr/24
Commented by mr W last updated on 21/Apr/24
((EY)/(YB))×(x/2)×((y+z)/z)=1  ⇒(y/z)=(2/x)−1=((2−x)/x)  ((FZ)/(ZC))×(y/z)×((p+q)/q)=1  ((2−x)/x)×(1+(p/q))=1  ⇒(p/q)=(x/(2−x))−1=((2(x−1))/(2−x))  ((DX)/(XA))×(p/q)×((x+2)/2)=1  ((2(x−1))/(2−x))×((x+2)/2)=1  x^2 +2x−4=0  ⇒x=(√5)−1 ✓
$$\frac{{EY}}{{YB}}×\frac{{x}}{\mathrm{2}}×\frac{{y}+{z}}{{z}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{y}}{{z}}=\frac{\mathrm{2}}{{x}}−\mathrm{1}=\frac{\mathrm{2}−{x}}{{x}} \\ $$$$\frac{{FZ}}{{ZC}}×\frac{{y}}{{z}}×\frac{{p}+{q}}{{q}}=\mathrm{1} \\ $$$$\frac{\mathrm{2}−{x}}{{x}}×\left(\mathrm{1}+\frac{{p}}{{q}}\right)=\mathrm{1} \\ $$$$\Rightarrow\frac{{p}}{{q}}=\frac{{x}}{\mathrm{2}−{x}}−\mathrm{1}=\frac{\mathrm{2}\left({x}−\mathrm{1}\right)}{\mathrm{2}−{x}} \\ $$$$\frac{{DX}}{{XA}}×\frac{{p}}{{q}}×\frac{{x}+\mathrm{2}}{\mathrm{2}}=\mathrm{1} \\ $$$$\frac{\mathrm{2}\left({x}−\mathrm{1}\right)}{\mathrm{2}−{x}}×\frac{{x}+\mathrm{2}}{\mathrm{2}}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{5}}−\mathrm{1}\:\checkmark \\ $$
Answered by A5T last updated on 21/Apr/24
((BD)/(DC))=(k/2);((BC)/(DC))=((k+2)/2)  ((AF)/(FB))×((BC)/(CD))×((DX)/(AX))=1⇒((AF)/(FB))=(2/(k+2))⇒((FB)/(BA))=((k+2)/(4+k))  ((AE)/(EC))×((CZ)/(ZF))×((FB)/(BA))=1⇒((AE)/(EC))=((4+k)/(k+2))  ((BD)/(DC))×((CA)/(AE))×((EY)/(EB))=1⇒(((k)(6+2k))/(2(4+k)))=1  ⇒6k+2k^2 =8+2k⇒k^2 −2k−4=0⇒k=(√5)−1
$$\frac{{BD}}{{DC}}=\frac{{k}}{\mathrm{2}};\frac{{BC}}{{DC}}=\frac{{k}+\mathrm{2}}{\mathrm{2}} \\ $$$$\frac{{AF}}{{FB}}×\frac{{BC}}{{CD}}×\frac{{DX}}{{AX}}=\mathrm{1}\Rightarrow\frac{{AF}}{{FB}}=\frac{\mathrm{2}}{{k}+\mathrm{2}}\Rightarrow\frac{{FB}}{{BA}}=\frac{{k}+\mathrm{2}}{\mathrm{4}+{k}} \\ $$$$\frac{{AE}}{{EC}}×\frac{{CZ}}{{ZF}}×\frac{{FB}}{{BA}}=\mathrm{1}\Rightarrow\frac{{AE}}{{EC}}=\frac{\mathrm{4}+{k}}{{k}+\mathrm{2}} \\ $$$$\frac{{BD}}{{DC}}×\frac{{CA}}{{AE}}×\frac{{EY}}{{EB}}=\mathrm{1}\Rightarrow\frac{\left({k}\right)\left(\mathrm{6}+\mathrm{2}{k}\right)}{\mathrm{2}\left(\mathrm{4}+{k}\right)}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{6}{k}+\mathrm{2}{k}^{\mathrm{2}} =\mathrm{8}+\mathrm{2}{k}\Rightarrow{k}^{\mathrm{2}} −\mathrm{2}{k}−\mathrm{4}=\mathrm{0}\Rightarrow{k}=\sqrt{\mathrm{5}}−\mathrm{1} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *