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Question Number 108067 by I want to learn more last updated on 14/Aug/20

Answered by 1549442205PVT last updated on 14/Aug/20

Commented by 1549442205PVT last updated on 14/Aug/20

$$\mathrm{BD}=2\mathrm{a},\mathrm{CD}=3\mathrm{a},\mathrm{CE}=12,\mathrm{AE}=6,\mathrm{AF}=3$$$$\mathrm{BF}=\mathrm{x}?$$$$\mathrm{First}\mathrm{we}\mathrm{prove}\mathrm{following}{\mathrm{Xeva}}^{\prime }\mathrm{s}\mathrm{therem}$$$$‘‘\mathrm{Given}\mathrm{the}\mathrm{triangle}\mathrm{ABC}\mathrm{and}\mathrm{D},\mathrm{E},\mathrm{F}$$$$\mathrm{are}\mathrm{three}\mathrm{points}\mathrm{lying}\mathrm{on}\mathrm{the}\mathrm{sides}\mathrm{BC}$$$$\mathrm{AC}\mathrm{and}\mathrm{AB}\mathrm{respectively}.\mathrm{Then}\mathrm{three}$$$$\mathrm{lines}\mathrm{AD},\mathrm{BE}\mathrm{and}\mathrm{CF}\mathrm{are}\mathrm{concurren}\mathrm{at}$$$$\mathrm{a}\mathrm{point}\mathrm{G}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\frac{\mathrm{BD}}{\mathrm{DC}}.\frac{\mathrm{CE}}{\mathrm{EA}}.\frac{\mathrm{AF}}{\mathrm{FB}}=1{(\ast )}^{″}$$$$\mathrm{i})\mathrm{Let}\mathrm{AD},\mathrm{BE},\mathrm{CF}\mathrm{are}\mathrm{omcurrent}\mathrm{at}\mathrm{G}$$$$\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{prove}\mathrm{the}\mathrm{equality}(\ast )$$$$\mathrm{Indeed},\mathrm{we}\mathrm{have}\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{{\mathrm{S}}_{\mathrm{ABD}}}{{\mathrm{S}}_{\mathrm{ACD}}}=\frac{{\mathrm{S}}_{\mathrm{GBD}}}{{\mathrm{S}}_{\mathrm{GCD}}}=$$$$=\frac{{\mathrm{S}}_{\mathrm{ABD}}-{\mathrm{S}}_{\mathrm{GBD}}}{{\mathrm{S}}_{\mathrm{ACD}}-{\mathrm{S}}_{\mathrm{GCD}}}=\frac{{\mathrm{S}}_{\mathrm{AGB}}}{{\mathrm{S}}_{\mathrm{AGC}}}\left(1\right).\mathrm{Similarly},\mathrm{we}$$$$\mathrm{have}\mathrm{also}\frac{\mathrm{CE}}{\mathrm{EA}}=\frac{{\mathrm{S}}_{\mathrm{BGC}}}{{\mathrm{S}}_{\mathrm{BGA}}}\left(2\right),\frac{\mathrm{AF}}{\mathrm{FB}}=\frac{{\mathrm{S}}_{\mathrm{AGC}}}{{\mathrm{S}}_{\mathrm{BGC}}}\left(3\right)$$$$\mathrm{Multiplying}\mathrm{three}\mathrm{above}\mathrm{equalities}$$$$\mathrm{together}\mathrm{we}\mathrm{obtain}$$$$\frac{\mathrm{BD}}{\mathrm{DC}}.\frac{\mathrm{CE}}{\mathrm{EA}}.\frac{\mathrm{AF}}{\mathrm{FB}}=\frac{{\mathrm{S}}_{\mathrm{AGB}}}{{\mathrm{S}}_{\mathrm{AGC}}}\times \frac{{\mathrm{S}}_{\mathrm{BGC}}}{{\mathrm{S}}_{\mathrm{BGA}}}\times \frac{{\mathrm{S}}_{\mathrm{AGC}}}{{\mathrm{S}}_{\mathrm{BGC}}}=1(\mathrm{q}.\mathrm{e}.\mathrm{d})$$$$\mathrm{Now}\mathrm{suppose}\mathrm{had}\mathrm{the}\mathrm{equality}(\ast )\mathrm{we}$$$$\mathrm{prove}\mathrm{AD},\mathrm{BE},\mathrm{CF}\mathrm{are}\mathrm{concurrent}.\mathrm{Indeed},$$$$\mathrm{denote}\mathrm{by}\mathrm{G}\mathrm{the}\mathrm{intersection}\mathrm{point}$$$$\mathrm{of}\mathrm{BE}\mathrm{and}\mathrm{CF}.\mathrm{Let}\mathrm{the}\mathrm{ray}\mathrm{AG}\mathrm{meet}\mathrm{BC}$$$$\mathrm{at}\mathrm{the}\mathrm{point}{\mathrm{D}}^{\prime }.\mathrm{Then}\mathrm{by}\mathrm{above}\mathrm{result}$$$$\mathrm{we}\mathrm{get}\frac{{\mathrm{BD}}^{\prime }}{{\mathrm{D}}^{\prime }\mathrm{C}}.\frac{\mathrm{CE}}{.\mathrm{EA}}.\frac{\mathrm{AF}}{\mathrm{FB}}=1.\mathrm{Then}\mathrm{combining}$$$$\mathrm{to}(\ast )\mathrm{we}\mathrm{get}\frac{{\mathrm{BD}}^{\prime }}{{\mathrm{D}}^{\prime }\mathrm{C}}=\frac{\mathrm{BD}}{\mathrm{DC}}\iff \frac{{\mathrm{BD}}^{\prime }}{\mathrm{BC}}=\frac{\mathrm{BD}}{\mathrm{BC}}$$$$\Rightarrow {\mathrm{D}}^{\prime }\equiv \mathrm{D}\Rightarrow \mathrm{AD}\mathrm{which}\mathrm{shows}\mathrm{that}\mathrm{AD},$$$$\mathrm{BE}\mathrm{and}\mathrm{CF}\mathrm{are}\mathrm{concurrent}\mathrm{at}\mathrm{G}(\mathrm{q}.\mathrm{e}.\mathrm{d})$$$$\mathrm{Now}\mathrm{applying}{\mathrm{Xeva}}^{\prime }\mathrm{s}\mathrm{theorem}\mathrm{on}\mathrm{the}$$$$\mathrm{given}\mathrm{problem}\mathrm{we}\mathrm{get}$$$$\frac{\mathrm{BD}}{\mathrm{DC}}.\frac{\mathrm{CE}}{\mathrm{EA}}.\frac{\mathrm{AF}}{\mathrm{FB}}=1\iff \frac{2\mathrm{a}}{3\mathrm{a}}.\frac{12}{6}.\frac{3}{\mathrm{x}}=1\Rightarrow \frac{72\mathrm{a}}{18\mathrm{ax}}=1$$$$\Rightarrow \mathrm{x}=4$$

Commented by I want to learn more last updated on 14/Aug/20

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}$$

Answered by floor(10²Eta[1]) last updated on 14/Aug/20

$$\mathrm{by}{\mathrm{ceva}}^{\prime }\mathrm{s}\mathrm{theorem}:$$$$\frac{2\mathrm{a}}{3\mathrm{a}}\times \frac{12}{6}\times \frac{3}{\mathrm{x}}=1\Rightarrow \mathrm{x}=4$$

Commented by I want to learn more last updated on 14/Aug/20

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}$$

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