Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

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

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

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

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!

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

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

thank you Sir

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com