A parallelogram ABCD has P the mid-point of DC and Q a point of AC such that CQ = $\dfrac{1}{4}AC$. PQ produced meets BC at R.

Prove that :

(i) R is the mid-point of BC,

(ii) PR = $\dfrac{1}{2}DB$.

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 of a triangle parallel to another, bisects the third side.

We know that,

Diagonals of parallelogram bisect each other.

∴ AX = CX and BX = DX.

Given,

⇒ CQ = $\dfrac{1}{4}AC$

⇒ CQ = $\dfrac{1}{4} \times (AX + CX)$

⇒ CQ = $\dfrac{1}{4} \times (CX + CX) = \dfrac{1}{4} \times 2CX = \dfrac{CX}{2}$.

∴ Q is the mid-point of CX.

(i) In △ CDX,

P and Q are mid-point of sides CD and CX.

∴ PQ || DX (By mid-point theorem)

Since, DXB and PQR are straight line.

∴ PR || DB

∴ QR || XB.

In △ CXB,

Q is the mid-point of CX and QR || XB.

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

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

(ii) In △ BCD,

P and R are mid-point of sides CD and BC.

∴ PR || BD and PR = $\dfrac{1}{2}BD$.

Hence, proved that PR = $\dfrac{1}{2}BD$.