Question

Prove that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer 1

Hint: First draw the diagram of a parallelogram ABCD with its diagonal as AC and BD. To prove that the area of two triangles is equal, use the theorem that triangles on the same base and between the same two parallel lines have equal area and also use the theorem that the median divides the triangle into two equal area triangles.

Complete step-by-step answer:
In the question, we have a parallelogram and we have to prove that the diagonals divide the parallelogram into four triangles of equal area.
The diagram for the question is given as:

From the diagram we can say that we to prove that:
${\text{ar(}}\vartriangle {\text{AOB) = ar(}}\vartriangle {\text{BOC) = ar(}}\vartriangle {\text{COD) = ar(}}\vartriangle {\text{DOA)}}$
We know the theorem that triangles on the same base and between the same two parallel lines have equal area.
Now, In $\vartriangle {\text{ABC and }}\vartriangle {\text{BCD}}$
BC is the common side i.e. they are on the same base.
And also they are between the same parallel lines AD and BC.
Therefore, ${\text{ar(}}\vartriangle {\text{ABC) = ar(}}\vartriangle {\text{BCD)}}$ (1)

Equation 1 can be written as:
\[\;{\text{ar(}}\vartriangle {\text{AOB) + ar(BOC) = ar(}}\vartriangle {\text{COD) + ar(}}\vartriangle {\text{BOC)}}\]
\[ \Rightarrow \;{\text{ar(}}\vartriangle {\text{AOB) = ar(}}\vartriangle {\text{COD)}}\] (2)
Now, we know that the diagonals of a parallelogram bisect each other.
$\therefore $ OA=OC
Or we can say that BO is the median of $\vartriangle {\text{ABC}}$ .
According to theorem on median of a triangle:
 The median of a triangle divides the triangle in two triangles of equal area.
So, using this theorem we can write:
${\text{ar(}}\vartriangle {\text{AOB) = ar(}}\vartriangle {\text{BOC)}}$ (3)
In $\vartriangle {\text{ACD}}$ , OD is the median.
Therefore, we can write:
${\text{ar(}}\vartriangle {\text{AOD) = ar(}}\vartriangle {\text{COD)}}$ (4)
From equation 1, 2, 3 and 4, we can write:
${\text{ar(}}\vartriangle {\text{AOB) = ar(}}\vartriangle {\text{BOC) = ar(}}\vartriangle {\text{COD) = ar(}}\vartriangle {\text{DOA)}}$

Note: In this type of question, we should remember the theorem on the area of a triangle. There are two theorems on the area of triangles. First one is that triangles on the same base and between same parallel lines have equal area and second one is that the median of the triangle divides the triangle into two triangles of equal area.