Question

Show that the diagonals of a || gm divide into four triangles of equal area.
Figure

Answer 1

Given: ABCD is a parallelogram. Its diagonals AC and BD intersect at O.
To prove:
ar(∆OAB) = ​ar(∆OBC) = ​ar(∆OCD) = ​ar (∆OAD)

Proof: Since the diagonals of a ​parallelogram bisects each other, we have:
OA = OC and OB = OD
Also, a median of a triangle divides it into two triangles of equal areas.
Now, in ​∆ABC, BO is the median.
∴ ar(​∆OAB) = ar(​∆OBC) ...(i)
In ​∆ABD, AO is the median.
∴ ar(∆OAB) = ar(∆OAD) ...(ii)

Now, in ∆ ADC, DO is the median.
∴ ar(∆OCD) = ar(∆OAD) ...(iii)
Similarly, in ∆BDC, OC is the median.
∴ ar(∆OBC) = ar(∆ODC) ...(iv)
From (i), (ii), (iii) and (iv), we get:
ar(∆OAB) = ​ar(∆OBC) = ​ar(∆OCD) = ​ar(∆OAD)
Hence, the diagonals of a parallelogram divide it into four triangles of equal areas.