Question

O is a point inside a triangle ABC. The Bisectors Of Angle Of AOB, Angle BOS And Angle COA meet the sides BC , CA AND AB in Point AB In Point E , F and D Respectively.Prove That AD.BE.CEF = DB.EC.FA​

Answer 1

$\sf\large{\underline{\underline{In\:Triangle\:AOD,}}}$

$\sf\large{OD\:is\:the\:bisector\:of\:angle\:AOB}$

$\sf\large{\red{\implies \dfrac{OA}{OB} = \dfrac{AD}{DB}\: \: ........(i)}}$

$\sf\large{ \: \: }$

$\sf\large{\underline{\underline{In\:Triangle\:BOC,\:OE\:is\: the\:Bisector\:of\:angle\:BOC.}}}$

$\sf\large{\purple{\implies \dfrac{OB}{OC} = \dfrac{BE}{EC} \: \: ........(ii)}}$

$\sf\large{ \: \: }$

$\sf\large{\underline{\underline{In\:Triangle\:COA}}}$

$\sf\large{OF\:is \: the\:bisector\:of\:angle\:COA.}$

$\sf\large{\orange{\implies \dfrac{OC}{OA} = \dfrac{CF}{FA} \: \: ......(iii)}}$

$\sf\large{ \: \: }$

$\sf\large{\longrightarrow Multiplying\:the\:corresponding\:sides\:of\:(i),\:(ii)\:and\:(iii)}$

$\sf\large{We\:get}$

$\sf\large{\red{\implies \dfrac{OA}{OB} × \dfrac{OB}{OC} × \dfrac{OC}{OA} = \dfrac{AD}{DB} × \dfrac{BE}{EC} × \dfrac{CF}{FA}}}$

$\sf\large{ \: \: }$

$\sf\large{\purple{\implies 1 = \dfrac{AD}{DB} × \dfrac{BE}{EC} × \dfrac{CF}{FA}}}$

$\sf\large{ \: \: }$

$\sf\large{\orange{\implies DB× EC × FA = AD × BE × CF}}$

$\sf\large{ \: \: }$

$\sf\large{\underbrace{\green{\implies AD × BE × CF = DB × EC × FA}}}$

$\sf\large{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$$\sf\large \: {\bold{\fbox{\color{red}{Hence\:Proved}}}}$