Question

In parallelogram ABCD point K belongs to diagonal

Answer 1

Answer:

$\dfrac{BE}{EC}=\dfrac{1}{3}.$

Step-by-step explanation:

Consider triangles BKE and DKA. In this triangles:

• $\angle BKE=\angle DKA$ as vertical angles;
• $\angle KBE=\angle KDA$ as alternate interior angles (lines BC and AD are parallel and BC is a transversal);
• $\angle BEK=\angle DAK$ 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:

$\dfrac{BK}{KD}=\dfrac{BE}{AD}\Rightarrow \dfrac{1}{4}=\dfrac{BE}{AD}.$

Thus, $AD=4BE.$

Since AD=BC and BC=BE+CE, we have that 4BE=BE+EC, EC=3BE. Hence, the ratio BE to EC is

$\dfrac{BE}{EC}=\dfrac{BE}{3BE}=\dfrac{1}{3}.$