The-median-AD-of-the-triangle-ABC-is-bisected-at-E-BE-meets-AC-in-F-then-AF-AC-

Question Number 54422 by gunawan last updated on 03/Feb/19

$$\mathrm{The}\:\mathrm{median}\:{AD}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:{ABC} \\ $$$$\mathrm{is}\:\mathrm{bisected}\:\mathrm{at}\:{E},\:{BE}\:\mathrm{meets}\:{AC}\:\mathrm{in}\:{F}, \\ $$$$\mathrm{then}\:{AF}\::\:{AC}\:= \\ $$

Commented by ajfour last updated on 03/Feb/19

Commented by ajfour last updated on 03/Feb/19

let BC = a^� , BA=c^� , AF =λ(a^� −c^� ) eq. of BF : r_F ^� =μ(c^� +(a^� /2))=c^� +λ(a^� −c^� ) ⇒ μ=1−λ and (μ/2)=λ ⇒ 2λ=1−λ hence λ= (1/3) or ((AF)/(AC)) = (1/3) .

$${let}\:{BC}\:=\:\bar {{a}}\:,\:{BA}=\bar {{c}}\:,\:{AF}\:=\lambda\left(\bar {{a}}−\bar {{c}}\right) \\ $$$${eq}.\:{of}\:{BF}\::\:\:\:\bar {{r}}_{{F}} =\mu\left(\bar {{c}}+\frac{\bar {{a}}}{\mathrm{2}}\right)=\bar {{c}}+\lambda\left(\bar {{a}}−\bar {{c}}\right) \\ $$$$\Rightarrow\:\mu=\mathrm{1}−\lambda\:\:\:{and}\:\:\frac{\mu}{\mathrm{2}}=\lambda \\ $$$$\Rightarrow\:\:\mathrm{2}\lambda=\mathrm{1}−\lambda\:\:\: \\ $$$${hence}\:\:\lambda=\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:{or}\:\:\:\frac{{AF}}{{AC}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

taking the help of drawing of sir Ajfour alternative proof... draw DK∣∣ CF ((BD)/(DC))=((BK)/(KF)) [since DK∣∣CF] ((BD)/(BC))=((BK)/(BF))=(1/2) so KD=(1/2)FC △DKE and △AEF <DEK=<AEF [vertically opposite angle] AE=ED given <KDE=<EAF[alternate angle] △s are congurent... so KD=AF=x (as per picture) AC=AF+FC=x+2x=3x so ((AF)/(AC))=(x/(3x))=(1/3)

$${taking}\:{the}\:{help}\:{of}\:{drawing}\:{of}\:{sir}\:{Ajfour} \\ $$$${alternative}\:{proof}… \\ $$$${draw}\:{DK}\mid\mid\:{CF} \\ $$$$\frac{{BD}}{{DC}}=\frac{{BK}}{{KF}}\:\left[{since}\:{DK}\mid\mid{CF}\right] \\ $$$$\frac{{BD}}{{BC}}=\frac{{BK}}{{BF}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{so}\:{KD}=\frac{\mathrm{1}}{\mathrm{2}}{FC} \\ $$$$\bigtriangleup{DKE}\:{and}\:\bigtriangleup{AEF} \\ $$$$<{DEK}=<{AEF}\:\left[{vertically}\:{opposite}\:{angle}\right] \\ $$$${AE}={ED}\:{given} \\ $$$$<{KDE}=<{EAF}\left[{alternate}\:{angle}\right] \\ $$$$\bigtriangleup{s}\:{are}\:{congurent}… \\ $$$${so}\:{KD}={AF}={x}\:\left({as}\:{per}\:{picture}\right) \\ $$$${AC}={AF}+{FC}={x}+\mathrm{2}{x}=\mathrm{3}{x} \\ $$$${so}\:\frac{{AF}}{{AC}}=\frac{{x}}{\mathrm{3}{x}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by mr W last updated on 03/Feb/19

Commented by ajfour last updated on 03/Feb/19

$${wow}!\:{man}\:{thinks},\:{God}\:{laughs}! \\ $$

Commented by mr W last updated on 03/Feb/19

draw DG//BF ∵ AE=ED and EF//DG ∴ AF=FG ∵DG//BF and BD=DC ∴ FG=GC ∵AF=FG=GC ∴ AF/AC=1/3

$${draw}\:{DG}//{BF} \\ $$$$\because\:{AE}={ED}\:{and}\:{EF}//{DG} \\ $$$$\therefore\:{AF}={FG} \\ $$$$ \\ $$$$\because{DG}//{BF}\:{and}\:{BD}={DC} \\ $$$$\therefore\:{FG}={GC} \\ $$$$ \\ $$$$\because{AF}={FG}={GC} \\ $$$$\therefore\:{AF}/{AC}=\mathrm{1}/\mathrm{3} \\ $$

Commented by gunawan last updated on 05/Feb/19

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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