Question

In the figure ABCD is a parallelogram and E divides BC in the ratio 1:3. DB and AE intersect at F. Show that DF=4FB and AF=4FE.

Answer 1

AE = a +1/4d
AF = ka + 1/4 kd

AF = AD + DF
= AD + hDB
= d + h(a-d)
= ha + (1-h)d

AF = AF
So,
ka + 1/4kd = ha + (1-h)d

From which k = h
And 1/4k = 1-h
since h=k

1/4k =1-k
Giving k = 4/5 and therefore, h = 4/5

DF = hDB
DF = 4/5DB
DF : FB = 4:1

∴ DF = 4FB

Also
AF = kAE
AF : FE = 4:1
∴ AF = 4FE

See the diagram below

Thanks

Answer 2

in the figure abcd is a parallelogram and E divides BC in the ratio 1:3 .DB and AE intersect at F