Consider a triangle $A B C$. The sides $A B$ and $A C$ are extended to points $D$ and $E,$ respectively, such that $A D=3 A B$ and $A B=3 A C$. Then one diagonal of $B D E C$ divides the other diagonal in the ratio ?
My approach
I am trying to find the relatio between BOC and and DOE (O is the intersecting point of the diagonals). But I am unable to do so. Although the ratio of AD/AB = AE/AC , I can't show $\triangle ABC$ and $\triangle ADE$ are equivalent. Am i going wrong?
Presumably you mean that $AE=3AC.$ If that is the case, then $\triangle ABC\sim \triangle ADE,$ so $BC$ is parallel to $DE.$ By the theorem on a transversal intersecting parallel lines, $\angle BCD=\angle CDE$ and $\angle CBE=\angle BED.$ So $$\triangle BOC\sim \triangle EOD.$$ By similarity ratios, $$\frac{BO}{EO}=\frac{CO}{DO}.$$ Is that what you were seeking?