Question

Page No:
Date: 1
CH- 6 Triangle
In a barallelogram A B C D a line. AE is drown on side BC such that BE: EC = 1:3 DB and AE intersects at F. Show that DF=4FB and AF=4AF.​

Answer 1

Given: The diagonal BO of parallelogarn ABCD intersects the segment AE at F,

where E is any point on BC.

To provo: DF x EF= FB x FA

Proof: In triangles AFD and BFE,

∠FAD = ∠FEB (Alternate angles)

∠AFD = ∠BFE (Vertically opposite angles)

Therefore △ADF ~ △BFE (AA similarity)

DF/FA = FB/EF

Hence DF x EF = FB x FA

so

I hope you are understand my solution