OC-4-5-amp-EF-2-amp-DE-EC-amp-AC-2AF-AB-

Question Number 223286 by mehdee7396 last updated on 20/Jul/25

$${OC}=\mathrm{4}.\mathrm{5}\:\:\:\&\:\:{EF}=\mathrm{2}\:\:\&\:\:{D}\overset{\frown} {{E}}={E}\overset{\frown} {{C}}\:\:\&\:\:{AC}=\mathrm{2}{AF}\Rightarrow{AB}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Downarrow\Downarrow\Downarrow \\ $$

Commented by fantastic last updated on 20/Jul/25

$$\mathrm{13}?? \\ $$$${Please}\:{tell}\:{me} \\ $$

Commented by mehdee7396 last updated on 20/Jul/25

Commented by fantastic last updated on 20/Jul/25

Answered by mr W last updated on 20/Jul/25

Commented by mr W last updated on 21/Jul/25

$${DE}={EC} \\ $$$$\Rightarrow{OE}\bot{CD} \\ $$$$\cancel{\Rightarrow\angle{COE}=\angle{EOD}=\angle{DOB}=\mathrm{60}°} \\ $$$${AF}={FC} \\ $$$$\Rightarrow{O},{E},\:{F}\:{are}\:{collinear}. \\ $$$${OF}=\mathrm{4}.\mathrm{5}+\mathrm{2}=\mathrm{6}.\mathrm{5} \\ $$$$\Rightarrow{AB}=\mathrm{2}×{OF}=\mathrm{2}×\mathrm{6}.\mathrm{5}=\mathrm{13}\:\checkmark \\ $$

Commented by mehdee7396 last updated on 21/Jul/25

$${in}\:{the}\:\mathrm{3}^{{th}} \:{line}\:{why}\:\:? \\ $$

Commented by A5T last updated on 21/Jul/25

$$\mathrm{I}\:\mathrm{suppose}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{actually}\:\mathrm{deduce}\: \\ $$$$\angle\mathrm{COE}=\angle\mathrm{EOD}=\angle\mathrm{DOB}=\mathrm{60}°\:\mathrm{just}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{information}\:\mathrm{DE}=\mathrm{EC}\:\mathrm{and}/\mathrm{or}\:\mathrm{OE}\bot\mathrm{CD}.\: \\ $$$$\mathrm{We}\:\mathrm{actually}\:\mathrm{need}\:\mathrm{more}\:\mathrm{information}\:\mathrm{to}\:\mathrm{deduce}\: \\ $$$$\mathrm{that}. \\ $$

Commented by fantastic last updated on 21/Jul/25

$${I}\:{will}\:\:{try}\:{to}\:{explain}\:{after}\:{some}\:{time} \\ $$$$\left[{I}\:{am}\:{also}\:{solving}\right] \\ $$

Commented by A5T last updated on 21/Jul/25

$$\mathrm{The}\:\mathrm{exact}\:\mathrm{value}\:\mathrm{of}\:\mathrm{any}\:\mathrm{angle}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{necessity}\:\mathrm{to} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{question}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{below}. \\ $$$$\mathrm{In}\:\mathrm{fact},\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{deduce}\:\angle\mathrm{COE}=\angle\mathrm{EOD}, \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{necessarily}\:\mathrm{unique}.\:\mathrm{And}\:\mathrm{AB}=\mathrm{2OF} \\ $$$$\mathrm{will}\:\mathrm{always}\:\mathrm{be}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\angle\mathrm{COE}=\angle\mathrm{EOD}=\theta. \\ $$

Commented by mr W last updated on 21/Jul/25

$${i}\:{didn}'{t}\:{use}\:{line}\:\mathrm{3}.\:{it}\:{is}\:{not}\:{true}\:{and} \\ $$$${not}\:{necessary}. \\ $$

Answered by A5T last updated on 21/Jul/25

$$\mathrm{OF}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{AB},\:\mathrm{since}\:\frac{\mathrm{CF}}{\mathrm{CA}}=\frac{\mathrm{CO}}{\mathrm{CB}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\angle\mathrm{COF}=\angle\mathrm{CBD} \\ $$$$\mathrm{Since}\:\mathrm{DE}=\mathrm{EC};\:\angle\mathrm{CBE}=\frac{\angle\mathrm{CBD}}{\mathrm{2}} \\ $$$$\angle\mathrm{COE}=\mathrm{2}\angle\mathrm{CBE}=\angle\mathrm{CBD} \\ $$$$\Rightarrow\angle\mathrm{COE}=\angle\mathrm{COF}\:\Rightarrow\mathrm{O},\mathrm{E},\mathrm{F}\:\mathrm{are}\:\mathrm{collinear}. \\ $$$$\frac{\mathrm{OF}}{\mathrm{AB}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{AB}=\mathrm{2OF}=\mathrm{2}\left(\mathrm{4}.\mathrm{5}+\mathrm{2}\right)=\mathrm{13} \\ $$

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