P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and Ris the mid-point of AP, show that
(i) ar (PRQ) =
ar (ARC)
(ii) ar (RQC) =
ar (ABC)
(iii) ar (PBQ) = ar (ARC)
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)