D and E are the midpoints of AB and AC.
DE will be parallel to BC (Midpoint theorem).
Area of ∆BEC and ∆BDC will be equal as they are on the same base and between  the same parallel lines.
Also, Area of ∆BEC - Area of ∆BOC = Area of ∆BDC - Area of ∆BOC.
Thus, Area of ∆DOB = Area of ∆EOC.

Answer: Option C. ->

It can be observed that parallelogram PBCS and PQRS are lying on the same base PS. However, these do not lie between the same parallel lines.

It can be observed that parallelogram ABCD and triangle PQR are lying between same parallel lines AD and BC. However, these do not have any common base.

It can be observed that parallelogram PQRS and triangle TQR have a common base QR and they are lying between the same parallel lines PS and QR.

It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS. However, their vertices, (i.e., opposite to the common base) P, Q of parallelogram and M, N of trapezium, are not lying on the same line.

CF when drawn parallel to AB, forms parallelogram ABCF.

AC is the diagonal of the parallelogram that divides ABCF into 2 equal triangles ABC and ACF.

So, Area of ΔACF = Area of ΔABC = $\frac{1}{2}\times 6\times 8=24c{m}^2$

Draw a line joining D and B.

See that parallelogram PDBQ and triangle ADB are on same base and between same parallels PQ and DB.

Area(ADB) = $\frac{1}{2}$ Area(PDBQ) = $\frac{1}{4}$ Area(PQRS)

Similarly, Area(DCB) = $\frac{1}{2}$ Area(DBRS) = $\frac{1}{4}$ Area(PQRS)

Area(DABC) = Area(ADB) + Area(DCB) = $\frac{1}{2}$ Area(PQRS)

Area(PQRS) = 2 × Area(DABC)

Consider the figure below:

In $△DAO\text{and}△OCB$
$\angle DAO=\angle OCB\text{(Alternate angles)}\angle ADO=\angle CBO\text{(Alternate angles)}AD=CB\text{(Opposite sides of parallelogram)}\text{Hence,}\Delta ADO\cong \Delta CBO\text{(AAS congruence)}\text{Similarly it can be proved that}Area\left(\Delta DOC\right)=Area\left(\Delta \right(BOA)$
In $△ABD$,  AO is the median.
A median divides a triangle into two equal parts. 
Hence 1 = 4
Similarly 2 = 3
$\therefore 1=2=3=4$​

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

So, Area $△$ BEC =  $\frac{1}{2}$ × Area parallelogram ABCD

Area $△$ BEC = $\frac{1}{2}$ × 1728

                      = 864 square units

Parallelogram ABCD and triangle ABD are on the same base and between same parallels .

Area($△ADB$) = $\frac{1}{2}$ Area of parallelogram(ABCD)

As P and Q are the midpoints of DC and AB, Area of parallelogram(AQPD) = $\frac{1}{2}$Area of parallelogram(ABCD)

So, Area($△$ADB) = Area of parallelogram(AQPD)

In fig 1, BC is the base for quadrilateral ABCD and ΔBEC lies between two parallel lines (BC and AD). In fig 4, AB is the base for quadrilateral ABCD and ΔABD lies between two parallel lines. Hence, options A and D are correct.

Given
BC = 6 cm
AD = BC (parallel sides of a parallelogram are equal)
Area (ABCD) = Base × Height
                        = AB × DF = AD × BE
AB × DF = 15 cm × 4 cm
                = 60 square cm
AD × BE = 60 square cm
$\Rightarrow$BE × 6 cm = 60 square cm
$\Rightarrow$BE = 10 cm

A median of the triangle divides it into two triangles of equal area.

So, Area ($△ABC$) = 2 × Area ($△ADB$) = 2 × 18 = 36 square cm.