Question Number 220511 by mehdee7396 last updated on 14/May/25
$${AB}=\mathrm{2}{CE}\:\:\&\:\:{DE}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}\:\:\: \\ $$$${CE}\:\bot{AB}\:\:\:\&\:\:\:{AD}\bot{BC}\:\:\&\:\:{AB}={AC}\:\:\&\:{EF}\:\bot{BC}\: \\ $$$${BF}=? \\ $$$$ \\ $$
Answered by mehdee7396 last updated on 14/May/25
Answered by mr W last updated on 14/May/25
Commented by mr W last updated on 15/May/25
$${yes} \\ $$
Commented by mr W last updated on 15/May/25
$${BD}={DC}={DE}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}={a},\:{say} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha=\frac{{CE}}{{AC}}=\frac{{CE}}{{AB}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${EF}={DE}\:\mathrm{sin}\:\mathrm{2}\alpha={a}\:\mathrm{sin}\:\mathrm{2}\alpha=\frac{{a}}{\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${BF}={BE}\:\mathrm{sin}\:\alpha=\mathrm{2}{a}\:\mathrm{sin}^{\mathrm{2}} \:\alpha={a}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{1} \\ $$
Commented by Ghisom last updated on 15/May/25
$$\mathrm{but}\:{BF}\:\mathrm{is}\:\mathrm{asked}\:\mathrm{for}\:\mathrm{and}\:\mathrm{I}\:\mathrm{get}\:{BF}=\mathrm{1} \\ $$
Commented by mehdee7396 last updated on 15/May/25
$${Bravo}\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by mehdee7396 last updated on 15/May/25
$$\bigtriangleup{ABC}\sim\bigtriangleup{BDE}\Rightarrow\frac{{EF}}{{EC}}=\frac{{ED}}{{AC}}\Rightarrow{EF}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{EF}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${BD}={ED}=\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\Rightarrow{BF}=\mathrm{1}\:\checkmark\: \\ $$$$ \\ $$