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:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

`(BK)/(CK) =(1)/(4)`

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK and Δ EBK are similar

so we have

`(AD)/(BE) =(DK)/(BK)`

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

`(BC)/(BE)=(4)/(1)` ( ABCD is parallelogram so AD=BC)

`(BE+EC)/(BE)=(4)/(1)` ( BC= BE+EC)

`(BE)/(BE) +(EC)/(BE)=(4)/(1)`

`1+(EC)/(BE)=4`

`(EC)/(BE)=3` ( subtracting 1 from both side )

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

taking reciprocal both side

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