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

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

Commented by ajfour last updated on 07/Aug/17

$$\overrightarrow{\mathrm{BF}}=(\mathrm{m}+1)\overline{\mathrm{b}}+\mathrm{n}\overline{\mathrm{a}}$$$$\overrightarrow{\mathrm{BG}}=\lambda \overrightarrow{\mathrm{BF}}=\lambda [(\mathrm{m}+1)\overline{\mathrm{b}}+\mathrm{n}\overline{\mathrm{a}}].$$

Commented by ajfour last updated on 07/Aug/17

Commented by ajfour last updated on 07/Aug/17

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

Commented by Tinkutara last updated on 07/Aug/17

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

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