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

BE:EC= 1/3

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

⇒
`(AD)/(BE)`=
`(DK)/(BK)`,

Since, AD= BC,therefore

`(AD)/(BE)`=
`(BC)/(BE)`=
`(4)/(1)`

Now, BC= BE+EC, ⇒
`(BE+EC)/(BE)`=
`(4)/(1)`

⇒1+
`(EC)/(BE) = (4)/(1)`

⇒
`(EC)/(BE)= 3`

Reciprocating on both the sides, we get

`(BE)/(BC) = (1)/(3)`