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Question Number 174763 by dragan91 last updated on 10/Aug/22

Answered by a.lgnaoui last updated on 11/Aug/22

∠ADF ((sin A)/(DF))=((sin F)/(AD))=((sin D)/(24)) (1) ∠EBF ((sin B)/x)=((sin F)/3)=((sin E)/8) (2) ∠CDE ((sin D)/3)=((sin C)/(DE))=((sin E)/(CD)) (3) (2) and (3) ((sin E)/8)=((sin F)/3) sin E=8((sin F)/3) ((sin E)/(CD))=((sin D)/3) sin E=((CDsin D)/3) ((8sin F)/3)=((CDsin D)/3) CD=((8sin F)/(sin D)) ∠ADF ((sin F)/(AD))=((sin D)/(24)) ⇒((sin F)/(sin D))=((AD)/(24)) CD=8((AD)/(24)) , CD=((AD)/3) ⇒ AC=4CD ((CD)/(CI))=((CE)/(BC))=(1/2)=((DE)/(BI)) (DE∥BI) GD∣∣BC ((FE)/(FD))=((FB)/(FG)) ⇒ (x/(FD))=(8/(12))=(2/3) H∈EF ( BH∣∣AC) ∡DCB=∡CBH ∡C+∡D=∡B+∡F CE=BE⇒DE=EH =3 ∡HBF=∡A CD=BH EF=2×3=6 ⇒ x=6

$$\angle{ADF} \\ $$$$\frac{\mathrm{sin}\:\mathrm{A}}{\mathrm{DF}}=\frac{\mathrm{sin}\:\mathrm{F}}{\mathrm{AD}}=\frac{\mathrm{sin}\:\mathrm{D}}{\mathrm{24}}\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\angle{EBF} \\ $$$$\frac{\mathrm{sin}\:\mathrm{B}}{\mathrm{x}}=\frac{\mathrm{sin}\:\mathrm{F}}{\mathrm{3}}=\frac{\mathrm{sin}\:\mathrm{E}}{\mathrm{8}}\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\angle{CDE} \\ $$$$\frac{\mathrm{sin}\:\mathrm{D}}{\mathrm{3}}=\frac{\mathrm{sin}\:\mathrm{C}}{\mathrm{DE}}=\frac{\mathrm{sin}\:\mathrm{E}}{\mathrm{CD}}\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{and}\:\left(\mathrm{3}\right) \\ $$$$\frac{\mathrm{sin}\:\mathrm{E}}{\mathrm{8}}=\frac{\mathrm{sin}\:\mathrm{F}}{\mathrm{3}}\:\:\:\mathrm{sin}\:\mathrm{E}=\mathrm{8}\frac{\mathrm{sin}\:\mathrm{F}}{\mathrm{3}} \\ $$$$\frac{\mathrm{sin}\:\mathrm{E}}{\mathrm{CD}}=\frac{\mathrm{sin}\:\mathrm{D}}{\mathrm{3}}\:\:\:\:\mathrm{sin}\:\mathrm{E}=\frac{\mathrm{CDsin}\:\mathrm{D}}{\mathrm{3}} \\ $$$$\frac{\mathrm{8sin}\:\mathrm{F}}{\mathrm{3}}=\frac{\mathrm{CDsin}\:\mathrm{D}}{\mathrm{3}}\:\:\:\mathrm{CD}=\frac{\mathrm{8sin}\:\mathrm{F}}{\mathrm{sin}\:\mathrm{D}} \\ $$$$\angle{ADF}\:\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{F}}{\mathrm{AD}}=\frac{\mathrm{sin}\:\mathrm{D}}{\mathrm{24}}\:\Rightarrow\frac{\mathrm{sin}\:\mathrm{F}}{\mathrm{sin}\:\mathrm{D}}=\frac{\mathrm{AD}}{\mathrm{24}} \\ $$$$\mathrm{CD}=\mathrm{8}\frac{\mathrm{AD}}{\mathrm{24}}\:\:\:\:,\:\:\mathrm{CD}=\frac{\mathrm{AD}}{\mathrm{3}}\:\:\Rightarrow\:\:\:\mathrm{AC}=\mathrm{4CD} \\ $$$$\frac{\mathrm{CD}}{\mathrm{CI}}=\frac{\mathrm{CE}}{\mathrm{BC}}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{DE}}{\mathrm{BI}}\:\:\left(\mathrm{DE}\parallel\mathrm{BI}\right) \\ $$$$\mathrm{GD}\mid\mid\mathrm{BC}\:\:\:\frac{\mathrm{FE}}{\mathrm{FD}}=\frac{\mathrm{FB}}{\mathrm{FG}}\:\:\Rightarrow\:\:\:\:\:\frac{\mathrm{x}}{\mathrm{FD}}=\frac{\mathrm{8}}{\mathrm{12}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{H}\in\mathrm{EF}\:\:\:\left(\:\:\mathrm{BH}\mid\mid\mathrm{AC}\right)\:\:\measuredangle{DCB}=\measuredangle{CBH} \\ $$$$\measuredangle{C}+\measuredangle{D}=\measuredangle{B}+\measuredangle{F}\:\:\:{CE}={BE}\Rightarrow{DE}={EH}\:\:=\mathrm{3}\:\:\:\:\measuredangle{HBF}=\measuredangle{A}\:\:{CD}={BH} \\ $$$${EF}=\mathrm{2}×\mathrm{3}=\mathrm{6}\:\:\Rightarrow\:\:\:\:\mathrm{x}=\mathrm{6} \\ $$

Commented by Tawa11 last updated on 11/Aug/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by a.lgnaoui last updated on 11/Aug/22

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