In the adjoining figure, ABCD is a parallelogram and E is the midpoint of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L. Prove that
(i) AF = 2 DC and (ii) DF = 2DL.

Explanation

EB|| DL and ED ||BL `rArr` EBLD is a ||gm.
`therefore BL=ED =(1)/(2)AD =(1)/(2)BC=CL.`
Now, in `triangle DCL ` and FBL, we have
CL= BL (proved), `angle DLC = angle FLB " " ("vert. opp. " angle )`
` and angle CDL = angle BFL " " ("alt. int. "angle).`
`therefore triangle DCL ~= triangle FBL. `
` therefore DC = BF and DL = FL. `
Now, `BF = DC = AB rArr 2 AB = 2DC rArr AF = 2DC. `
`DL = FL rArr DF = 2DL.`