On ∆ABC, BC is extended to E such that CE=BC/2. IF D is the mid point of AC and ED cuts AB at F, Find the value of AB/AF.
+26367 On \(\triangle ABC\), BC is extended to E such that \(CE=\dfrac{BC}{2}\).
IF D is the mid point of AC and ED cuts AB at F,
Find the value of \(\dfrac{AB}{AF}\).
Ceva's theorem
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{BC}{CE}*\dfrac{EG}{GA}*\dfrac{AF}{BF}} &=& \mathbf{1} \\\\ \dfrac{2}{1}*\dfrac{EG}{GA}*\dfrac{AF}{BF} &=& 1 \\\\ \dfrac{EG}{GA} &=& \dfrac{1}{2}* \dfrac{BF}{AF} \\\\ \mathbf{\dfrac{GA}{EG}} &=& \mathbf{2* \dfrac{AF}{BF}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{AD}{DC}} &=& \mathbf{\dfrac{GA}{EG}+\dfrac{AF}{BF}} \\\\ \dfrac{1}{1} &=& 2* \dfrac{AF}{BF}+\dfrac{AF}{BF} \\\\ 1 &=& 3* \dfrac{AF}{BF} \\\\ \dfrac{AF}{BF} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{BF}{AF}} &=& \mathbf{\dfrac{3}{1}} \\ \hline BF &=& \dfrac{3}{4}AB \\ AF &=& \dfrac{1}{4}AB \\\\ \dfrac{AB}{AF} &=& \dfrac{AB}{\dfrac{1}{4}AB } \\\\ \mathbf{\dfrac{AB}{AF}} &=& \mathbf{4} \\ \hline \end{array}\)
