In the adjoining figure, ABCD is a parallelogram in which E is the mid-point of DC and F is a point on AC such that CF = $\dfrac{1}{4}$ AC. If EF is produced to meet BC in G, prove that G is the mid-point of BC.

Join BD. Let BD intersect AC at point O.

We know that,

The diagonals of a parallelogram bisect each other.

AO = CO and OD = OB

Given,

⇒ CF = $\dfrac{1}{4}$ AC

⇒ CF = $\dfrac{1}{4}$ (AO + CO)

⇒ CF = $\dfrac{1}{4}$ (CO + CO)

⇒ CF = $\dfrac{1}{4}$ 2 CO

⇒ CF = $\dfrac{1}{2}$ CO

∴ F is the mid-point of CO.

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 equal to half of it.

In △COD,

Since, E and F are the mid-points of DC and OC respectively.

EF || OD

Since, EG and BD are straight lines. Thus, FG || OB

By converse of mid-point theorem,

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

In △COB,

Since, F is the mid-point of CO and FG || OB

∴ G is the mid-point of BC.

Hence, proved that G is the mid-point of BC.