In the quadrilateral given below, AB || DC, E and F are mid-points of AD and BD respectively. Prove that

(i) G is the mid-point of BC

(ii) EG = $\dfrac{1}{2}$(AB + DC).

(i) In △ABD,

E is mid-point of AD and F is mid-point of BD,

∴ EF || AB and EF = $\dfrac{1}{2}$AB …….(1)

Given,

AB || CD

Since, EF || AB and AB || CD

⇒ EF || CD

⇒ EG || CD.

Since, EG || CD we can say,

In △BCD,

⇒ FG || CD

Given, F is midpoint of BD and FG || CD

∴ G is the midpoint of BC. (By converse of mid-point theorem)

Hence, proved that G is the midpoint of BC.

(ii) In △BCD,

F and G are midpoint of BD and BC respectively,

FG = $\dfrac{1}{2}$CD ……….(2)

Adding eqn. (1) from part (i) and eqn (2) we get,

EF + FG = $\dfrac{1}{2}$AB + $\dfrac{1}{2}$CD

EG = $\dfrac{1}{2}$(AB + CD).

Hence, proved that EG = $\dfrac{1}{2}$(AB + CD).