Question

In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is the ratio of BE:EC?

Answer 1

Answer:

BE:EC= 1/3

Step-by-step explanation:

Given ABCD is a parallelogram, Point K is such that it belongs to diagonal BD so that BK:DK=1:4.

If we make an extension of AK which meets BS at point E, then using ΔDKA and ΔEKB, we have

∠DKA=∠EKB (Vertically opposite angles)

∠KDA=∠KBE (Alternate interior angles)

∠DAK=∠BEK (Alternate interior angles)

Thus by AAA similarity,ΔDKA≅ΔEKB

⇒$\frac{AD}{BE}$= $\frac{DK}{BK}$,

Since, AD= BC,therefore

$\frac{AD}{BE}$= $\frac{BC}{BE}$= $\frac{4}{1}$

Now, BC= BE+EC, ⇒$\frac{BE+EC}{BE}$= $\frac{4}{1}$

⇒1+$\frac{EC}{BE} = \frac{4}{1}$

⇒$\frac{EC}{BE}= 3$

Reciprocating on both the sides, we get

$\frac{BE}{BC} = \frac{1}{3}$