P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that (i) ar(PRQ) = $$\frac{1}{2}$$ ar(ARC) (ii) ar(PBQ) = ar(ARC) (iii) ar(RQC) = $$\frac{3}{8}$$ ar(ABC)

We have a $$\triangle$$ABC such that P is the mid-point of AB and Q is the mid-point of BC.
Also, R is the mid-point of AP. Let us join AQ, RQ, PC and PC.

(i) In $$\triangle$$ APQ , R is the mid point of AP(Given )
$$\therefore$$ RQ is the median of $$\triangle$$ APQ
$$\Rightarrow$$ ar(PRQ) = $$\frac{1}{2}$$ ar (APQ) ..... (1 )

In triangle ABQ ,P is the mid point of AB
$$\therefore$$ QP is the median of $$\triangle$$ ABQ
$$\Rightarrow$$ ar(APQ) = $$\frac{1}{2}$$ ar (ABQ) ..... (2 )

from (1) and (2) we get
$$\Rightarrow$$ ar(PRQ) = $$\frac{1}{2} × \frac{1}{2}$$ ar (ABQ)
$$\Rightarrow$$ ar(PRQ) = $$\frac{1}{4}$$ ar (ABQ) = $$\frac{1}{4} × \frac{1}{2}$$ ar (ABC)
[ $$\because$$ AQ is the Median of $$\triangle$$ ABC ]
$$\therefore$$ ar(PRQ) = $$\frac{1}{8}$$ × ar (ABC).....(3)

Now ar(ARC) = $$\frac{1}{2}$$ ×ar(APC)
[ $$\because$$ CR is a Median of $$\triangle APC$$ ]

$$\Rightarrow$$ ar(ARC) = $$\frac{1}{2} \frac{1}{2}$$ × ar(ABC)
[ $$\because$$ CP is a Median of $$\triangle ABC$$ ]
$$\therefore$$ ar(ARC) = $$\frac{1}{4}$$ × ar(ABC) .....(4)

Now from (3) we get
ar(PRQ) = $$\frac{1}{8}$$ (ABC)
ar(PRQ) = $$\frac{1}{2} \frac{1}{4}$$ ar(ABC)
ar(PRQ) = $$\frac{1}{2}$$ (ARC)

(iii) In triangle RBC , RQ is a Median
$$\therefore$$ ar(RQC) = ar(RBQ)
$$\Rightarrow$$ ar(RQC) = ar(PRQ) + ar(BPQ)
$$\Rightarrow$$ ar(RQC) = $$\frac{1}{8}$$× ar(ABC) + ar(BPQ)
$$\Rightarrow$$ ar(RQC) = $$\frac{1}{8}$$ × ar(ABC) + ar(BPQ)
$$\Rightarrow$$ ar(RQC) = $$\frac{1}{8} × ar(ABC) + \frac{1}{2}$$ ar(BPQ)
[$$\because$$ PQ is the Median of BPC ]
$$\Rightarrow$$ ar(RQC) = $$\frac{1}{8} × ar(ABC) + \frac{1}{2} \frac{1}{2}$$ ar(ABC)
[$$\because$$ CP is the Median of ABC ]
$$\Rightarrow$$ ar(RQC) = ( $$\frac{1}{4} + \frac{1}{8}$$ ar(ABC)

Thus , ar(RQC) = ( $$\frac{3}{8}$$ ar(ABC)

(ii) QP is the Median of $$\triangle$$ ABQ
$$\therefore$$ ar$$\triangle$$ PBQ = $$\frac{1}{2}$$ ar(ABQ)
= $$\frac{1}{2}× \frac{1}{2}$$ × ar(ABC)
= $$\frac{1}{4}$$ × ar(ABC) = ar(ARC )

From equation (4)
Thus ar(PBQ) = ar(ARC)