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Question Number 139649 by mnjuly1970 last updated on 30/Apr/21

Commented by mr W last updated on 30/Apr/21

((red)/(green))=(6/5)

$$\frac{{red}}{{green}}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$

Answered by mr W last updated on 30/Apr/21

DE//AB (∗) AB is tangent at F. (CE+2)×2=4^2 (∗∗) ⇒CE=6 ((DE)/(AB))=((CE)/(CB))=(6/(6+2))=(6/8) ⇒DE=((6×(6+4))/8)=((15)/2) ((AC)/(DC))=((CB)/(CE))=(8/6)=(4/3) AC=(4/3)×DC AD×AC=6^2 (∗∗) (AC−DC)×AC=6^2 ((4/3)×DC−DC)×(4/3)×DC=6^2 (4/9)×DC^2 =6^2 ⇒DC=((6×3)/2)=9 ⇒((DC)/(DE))=((9×2)/(15))=(6/5)

$${DE}//{AB}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\ast\right) \\ $$$${AB}\:{is}\:{tangent}\:{at}\:{F}. \\ $$$$ \\ $$$$\left({CE}+\mathrm{2}\right)×\mathrm{2}=\mathrm{4}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\left(\ast\ast\right) \\ $$$$\Rightarrow{CE}=\mathrm{6} \\ $$$$\frac{{DE}}{{AB}}=\frac{{CE}}{{CB}}=\frac{\mathrm{6}}{\mathrm{6}+\mathrm{2}}=\frac{\mathrm{6}}{\mathrm{8}} \\ $$$$\Rightarrow{DE}=\frac{\mathrm{6}×\left(\mathrm{6}+\mathrm{4}\right)}{\mathrm{8}}=\frac{\mathrm{15}}{\mathrm{2}} \\ $$$$\frac{{AC}}{{DC}}=\frac{{CB}}{{CE}}=\frac{\mathrm{8}}{\mathrm{6}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${AC}=\frac{\mathrm{4}}{\mathrm{3}}×{DC} \\ $$$${AD}×{AC}=\mathrm{6}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\ast\ast\right) \\ $$$$\left({AC}−{DC}\right)×{AC}=\mathrm{6}^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{3}}×{DC}−{DC}\right)×\frac{\mathrm{4}}{\mathrm{3}}×{DC}=\mathrm{6}^{\mathrm{2}} \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}×{DC}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} \\ $$$$\Rightarrow{DC}=\frac{\mathrm{6}×\mathrm{3}}{\mathrm{2}}=\mathrm{9} \\ $$$$\Rightarrow\frac{{DC}}{{DE}}=\frac{\mathrm{9}×\mathrm{2}}{\mathrm{15}}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$

Commented by mr W last updated on 30/Apr/21

Commented by mnjuly1970 last updated on 30/Apr/21

very nice thanks alot mr W.. grateful..

$${very}\:{nice}\:{thanks}\:{alot}\:{mr}\:{W}.. \\ $$$${grateful}.. \\ $$

Commented by mr W last updated on 30/Apr/21

Commented by mr W last updated on 30/Apr/21

proof for (∗) CG is the common tangent of both circles at point C. ∠CDE=∠GCE=α ∠CAB=∠GCB=α ⇒∠CDE=∠CAB ⇒DE//AB

$${proof}\:{for}\:\left(\ast\right) \\ $$$${CG}\:{is}\:{the}\:{common}\:{tangent}\:{of}\:{both} \\ $$$${circles}\:{at}\:{point}\:{C}. \\ $$$$\angle{CDE}=\angle{GCE}=\alpha \\ $$$$\angle{CAB}=\angle{GCB}=\alpha \\ $$$$\Rightarrow\angle{CDE}=\angle{CAB} \\ $$$$\Rightarrow{DE}//{AB} \\ $$

Commented by mr W last updated on 30/Apr/21

proof for (∗∗) see Q137946

$${proof}\:{for}\:\left(\ast\ast\right) \\ $$$${see}\:{Q}\mathrm{137946} \\ $$

Commented by mnjuly1970 last updated on 30/Apr/21

thanks alot...

$${thanks}\:\:{alot}... \\ $$

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