Take a point Z on AC. And extend PQ to T such that PQ = QT

Now,

Join TC, QZ, PZ and AQ

Hence,

In ΔABC, P and Q are the mid points of AB and BC respectively

So, by applying mid-point theorem, we get:

PQ‖AC and PQ = � AC

PQ‖AZ and PQ = AZ (Z is the mid-point of AC)

PQZA is a parallelogram

As we know that diagonals of a parallelogram bisect the parallelogram into two triangles of equal area

Area (ΔPAZ) = area (ΔZQP) = area (ΔPAQ) = Area (ΔZQA)

Similarly,

It can also be proved that quadrilaterals QZCT, PZQB and PZCQ are also parallelograms.

Hence,

Area (ΔPZQ) = Area (ΔCQZ)

Area (ΔQZC) = Area (ΔCTQ)

Area (ΔPAQ) = Area (ΔQBP)

Hence,

Area (ΔPAZ) = Area (ΔZQP)= Area (ΔPAQ) = Area (ΔZAQ) = Area(ΔZCQ) = Area (ΔCTQ) = Area (ΔPBQ) (1)

And,

Area (ΔABC) = Area(ΔPBQ) + Area(ΔPAZ) + Area (ΔPZQ) + Area (ΔQZC)

Area (ΔABC) = Area (ΔPBQ) + Area (ΔPBQ) + Area(ΔPBQ) + Area(ΔPBQ)

Area (ΔABC) = 4 Area (ΔPBQ)

Area (ΔPBQ) = Area (ΔABC) (2)

Now,

(i) Join PC.

In ΔPAQ, QR is median

Hence,

Area (ΔPRQ) = � Area (ΔPAQ)

=

In (ΔABC), P and Q are the mid points of AB and BC respectively

By applying mid-point theorem, we get:

PQ = 1/2 AC

AC = 2PQ

AC=PT

And,

PQ ‖ AC

PT ‖ AC

Therefore,

PACT is a parallelogram

Area (PACT) = Area (PACQ) + Area (ΔQTC)

=Area (PACQ) + Area (ΔQBP) [Using (1)]

Area (PACT) = Area (ΔABC) (4)

Area (ΔARC) = � Area (ΔPAC) (CR is median of ΔPAC)

= � * � Area (PACT)

=

= Area (ΔABC)

Area (ΔARC) = Area (ΔABC)

� Area (ΔARC) = Area (ΔPRQ) (from third equation) (5)

(i) Area(PACT) = Area(ΔPRQ) + Area(ΔARC) + Area(ΔQTC) + Area(ΔRQC)

Now,

Using (1), (2), (3), (4) and (5), we have:

Area (ΔABC) =

Area (ΔABC) = )

Area (ΔRQC) = Area (ΔABC)

Area (ΔRQC)

(i) In parallelogram PACT, we can find that:

Area (ΔARC) = 1/ Area (ΔPAC) (CR median of triangle PAC)

= � * � Area (ΔPACT) (PC diagonal of parallelogram PACT)

=

= Area (ΔABC)

= Area (ΔPBQ)