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Question Number 19238 by Tinkutara last updated on 07/Aug/17

Let ABCD be a parallelogram. Two points E and F are chosen on the sides BC and CD, respectively, such that ((EB)/(EC)) = m, and ((FC)/(FD)) = n. Lines AE and BF intersect at G. Prove that the ratio ((AG)/(GE)) = (((m + 1)(n + 1))/(mn)).

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{parallelogram}.\:\mathrm{Two} \\ $$$$\mathrm{points}\:{E}\:\mathrm{and}\:{F}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{on}\:\mathrm{the}\:\mathrm{sides} \\ $$$${BC}\:\mathrm{and}\:{CD},\:\mathrm{respectively},\:\mathrm{such}\:\mathrm{that} \\ $$$$\frac{{EB}}{{EC}}\:=\:{m},\:\mathrm{and}\:\frac{{FC}}{{FD}}\:=\:{n}.\:\mathrm{Lines}\:{AE}\:\mathrm{and}\:{BF} \\ $$$$\mathrm{intersect}\:\mathrm{at}\:{G}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{ratio} \\ $$$$\frac{{AG}}{{GE}}\:=\:\frac{\left({m}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{1}\right)}{{mn}}. \\ $$

Commented by ajfour last updated on 07/Aug/17

BF^(→) =(m+1)b^� +na^� BG^(→) =λBF^(→) =λ[(m+1)b^� +na^� ] .

$$\overset{\rightarrow} {\mathrm{BF}}=\left(\mathrm{m}+\mathrm{1}\right)\bar {\mathrm{b}}+\mathrm{n}\bar {\mathrm{a}} \\ $$$$\overset{\rightarrow} {\mathrm{BG}}=\lambda\overset{\rightarrow} {\mathrm{BF}}=\lambda\left[\left(\mathrm{m}+\mathrm{1}\right)\bar {\mathrm{b}}+\mathrm{n}\bar {\mathrm{a}}\right]\:. \\ $$

Commented by ajfour last updated on 07/Aug/17

Commented by ajfour last updated on 07/Aug/17

let B be origin. BG^(→) =λ[(m+1)b^� +na^� ] ....(i) Also BG^(→) =BE^(→) +EG^(→) Let EG^(→) =μAE^(→) ⇒ AG^(→) =(1−μ)AE^(→) ⇒ BG^(→) =mb^� +μ[(n+1)a^� −mb^� ] ..(ii) comparing coefficients of a^� and b^� in (i) and (ii): nλ=(n+1)μ ; (m+1)λ=m(1−μ) ⇒ ((m(1−μ))/((n+1)μ))= (((m+1)λ)/(nλ)) ((AG)/(GE))=((1−μ)/μ)= (((m+1)(n+1))/(mn)) .

$$\mathrm{let}\:\mathrm{B}\:\mathrm{be}\:\mathrm{origin}. \\ $$$$\overset{\rightarrow} {\mathrm{BG}}=\lambda\left[\left(\mathrm{m}+\mathrm{1}\right)\bar {\mathrm{b}}+\mathrm{n}\bar {\mathrm{a}}\right]\:\:\:....\left(\mathrm{i}\right) \\ $$$$\mathrm{Also}\:\overset{\rightarrow} {\mathrm{BG}}=\overset{\rightarrow} {\mathrm{BE}}+\overset{\rightarrow} {\mathrm{EG}}\: \\ $$$$\mathrm{Let}\:\overset{\rightarrow} {\mathrm{EG}}=\mu\overset{\rightarrow} {\mathrm{AE}}\:\Rightarrow\:\overset{\rightarrow} {\mathrm{AG}}=\left(\mathrm{1}−\mu\right)\overset{\rightarrow} {\mathrm{AE}} \\ $$$$\Rightarrow\:\overset{\rightarrow} {\mathrm{BG}}=\mathrm{m}\bar {\mathrm{b}}+\mu\left[\left(\mathrm{n}+\mathrm{1}\right)\bar {\mathrm{a}}−\mathrm{m}\bar {\mathrm{b}}\right]\:\:\:..\left(\mathrm{ii}\right) \\ $$$$\mathrm{comparing}\:\mathrm{coefficients}\:\mathrm{of}\:\bar {\mathrm{a}}\:\mathrm{and}\:\bar {\mathrm{b}} \\ $$$$\mathrm{in}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right): \\ $$$$\mathrm{n}\lambda=\left(\mathrm{n}+\mathrm{1}\right)\mu\:\:;\:\left(\mathrm{m}+\mathrm{1}\right)\lambda=\mathrm{m}\left(\mathrm{1}−\mu\right) \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{m}\left(\mathrm{1}−\mu\right)}{\left(\mathrm{n}+\mathrm{1}\right)\mu}=\:\frac{\left(\mathrm{m}+\mathrm{1}\right)\lambda}{\mathrm{n}\lambda} \\ $$$$\frac{\mathrm{AG}}{\mathrm{GE}}=\frac{\mathrm{1}−\mu}{\mu}=\:\frac{\left(\mathrm{m}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{mn}}\:. \\ $$

Commented by Tinkutara last updated on 07/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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