In the adjoining figure, ABCD is a parallelogram, E is the mid-point of CD and through D, a line is drawn parallel to EB to meet CB produced at G and intersecting AB at F. Prove that :

(i) AD = $\dfrac{1}{2}$ GC

(ii) DG = 2 EB

By mid-point theorem,

The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side side of a triangle parallel to another, bisects the third side.

Given, DF || EB

⇒ DG || EB

In △ DGC,

⇒ E is the mid-point of CD and DG || EB.

∴ B is the mid-point of GC. (By converse of mid-point theorem)

∴ BG = BC

Thus, BC = $\dfrac{1}{2}GC$

(i) We know that,

Opposite sides of a parallelogram are equal.

⇒ AD = BC

⇒ AD = $\dfrac{1}{2}$ GC

Hence, proved that AD = $\dfrac{1}{2}$ GC.

(ii) In △DGC,

Since, E and B are the mid-points of DC and GC respectively.

EB || DG

Thus, by mid-point theorem,

⇒ EB = $\dfrac{1}{2}$ DG

⇒ DG = 2 EB.

Hence, proved that DG = 2EB.