P and Q are respectively the mid-point of sides AB and BC of a triangle ABC and R is the mid-point of AP. Show that (i) ar(PRQ) = 1/2 ar(ARC) (ii) ar(RQC) = 3/8 ar(ABC) (iii) ar(PBQ) = ar(ARC)

Let ar(ABC) denote the area of triangle ABC.

(i) In triangle ABP, R is the mid-point of AP. Therefore, ar(BRP) = ar(BRQ) = 1/2 ar(ABP).
Since P is the mid-point of AB, ar(ABP) = 1/2 ar(ABC).
Therefore, ar(BRP) = ar(BRQ) = 1/4 ar(ABC).
In triangle ABC, Q is the mid-point of BC. Therefore, ar(ABQ) = ar(ACQ) = 1/2 ar(ABC).
In triangle ARC, R is the mid-point of AP and P is the mid-point of AB. Therefore, ar(ARP) = ar(PRC) = 1/2 ar(ARC).
Also, ar(ARC) = 1/4 ar(ABC).
Therefore, ar(PRQ) = ar(PRC) - ar(QRC) = 1/4 ar(ABC) - 1/8 ar(ABC) = 1/8 ar(ABC).
Since ar(ARC) = 1/4 ar(ABC), ar(PRQ) = 1/2 ar(ARC).

(ii) ar(RQC) = ar(RQA) + ar(AQC) - ar(RAQ)
Since Q is the midpoint of BC, ar(AQC) = 1/2 ar(ABC)
ar(ABQ) = 1/2 ar(ABC)
ar(RQC) = ar(AQC) - ar(ARQ)
ar(ABQ) = 1/2 ar(ABC)
ar(APQ) = 1/4 ar(ABQ) = 1/8 ar(ABC)
ar(ARP) = 1/2 ar(APQ) = 1/16 ar(ABC)
ar(RQA) = 1/2 ar(APQ) = 1/16 ar(ABC)
ar(RQC) = ar(AQC) - ar(RAQ) = 1/2 ar(ABC) - 1/16 ar(ABC) = 7/16 ar(ABC)
However, ar(RQC) = ar(AQC) - ar(ARQ) = 1/2 ar(ABC) - 1/8 ar(ABC) = 3/8 ar(ABC)

(iii) ar(PBQ) = 1/2 ar(ABC) - 1/8 ar(ABC) = 3/8 ar(ABC)
ar(ARC) = 1/4 ar(ABC)
Therefore, ar(PBQ) ≠ ar(ARC)