State true or false:In the figure, ABCD is a parallelogram and E divides BC in the ratio 1 : 3. DB and AE intersect at F then DF=4BF and AF=4FE.

It is given that ABCD is a parallelogram and E divides BC in the ratio

1 : 3

. DB and AE intersect at F.

Δ FAD

and

Δ FEB

∠ FAD = ∠ FEB

(Alternate interior angles)

∠ FDA = ∠ FBE

(Alternate interior angles)
So by AA similarity criterion,

Δ FAD \sim ΔFEB

We know that the corresponding sides of similar triangles have the same ratios, then

\frac{AF}{EF} = \frac{AD}{BE} = \frac{DF}{BF}

As opposite sides of parallelogram are equal,

AD = BC

then,

\frac{AF}{EF} = \frac{BC}{BE} = \frac{DF}{BF}

………….(1)
As

BE : EC = 1 : 3

\Rightarrow \frac{BE}{EC} = \frac{1}{3}

\Rightarrow EC = 3 BE

So,

BC = BE + CE

\Rightarrow BC = BE + 3 BE

\Rightarrow BC = 4 BE

Put the value of BC in (1).

\frac{AF}{EF} = \frac{4 BE}{BE} = \frac{DF}{BF}

\Rightarrow \frac{AF}{EF} = \frac{4}{1} = \frac{DF}{BF}

………..(2)
From (2) we get,

\frac{4}{1} = \frac{DF}{BF}

\Rightarrow DF = 4 BF

And,

\frac{AF}{EF} = \frac{4}{1}

\Rightarrow AF = 4 EF

Therefore, it is true that ABCD is a parallelogram and E divides BC in the ratio

1 : 3

. DB and AE intersect at F then

DF = 4 BF

and

AF = 4 FE

.
Hence, option 1 is correct.

State true or false:

In the figure, ABCD is a parallelogram and E divides BC in the ratio $1 : 3$ . DB and AE intersect at F then $DF = 4 BF$ and $AF = 4 FE$ .