How prove that $AD>BE$ in triangle?

Question

Let $D$ be a point on the side $BC$ of a triangle $ABC$ such that $AD>BC$ . The point $E$ on $CA$ is defined by the equation $\frac{AE}{EC}=\frac{BD}{AD-BC}$ .How prove that $AD>BE$?

Answer 1

This is quite subtle. Take $F$ on $AD$ such that $AF=BC$, and set $BD=x,DC=y,FD=z$.

Due to Menelaus theorem, $B,F$ and $E$ are collinear.

Take $G$ as the intersection between $CF$ and $AB$. Since $\frac{BD}{DC}=\frac{x}{y}$ and $\frac{CE}{EA}=\frac{z}{x}$, the converse of Ceva's theorem gives $\frac{AG}{GB}=\frac{y}{z}$, and Van Obel's theorem gives $\frac{BF}{FE}=\frac{x+z}{y}$. Since the triangular inequality ensures: $$BF\leq BD+DF = x+z,$$ we have: $$BE = \frac{x+y+z}{x+z}BF \leq x+y+z = AD$$ as wanted.