Answer:
[tex]\dfrac{BE}{EC}=\dfrac{1}{3}.[/tex]
Step-by-step explanation:
Consider triangles BKE and DKA. In this triangles:
- [tex]\angle BKE=\angle DKA[/tex] as vertical angles;
- [tex]\angle KBE=\angle KDA[/tex] as alternate interior angles (lines BC and AD are parallel and BC is a transversal);
- [tex]\angle BEK=\angle DAK[/tex] as alternate interior angles (lines BC and AD are parallel and BE is a transversal).
Thus triangles BKE and DKA are similar by AA theorem. Similar triangles have proportional sides lengths:
[tex]\dfrac{BK}{KD}=\dfrac{BE}{AD}\Rightarrow \dfrac{1}{4}=\dfrac{BE}{AD}.[/tex]
Thus, [tex]AD=4BE.[/tex]
Since AD=BC and BC=BE+CE, we have that 4BE=BE+EC, EC=3BE. Hence, the ratio BE to EC is
[tex]\dfrac{BE}{EC}=\dfrac{BE}{3BE}=\dfrac{1}{3}.[/tex]
