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

**Solution:**

We know that the diagonals of a parallelogram bisect each other.

Also, the median of a triangle divides it into two triangles of equal areas. By the use of these observations, we can get the required result.

Let's draw a diagram according to the question statement.

We know that diagonals of parallelograms bisect each other. Therefore, O is the mid-point of diagonal AC and BD.

BO is the median in Δ*ABC*. Therefore, BO will divide Δ*ABC* into two triangles of equal areas.

∴ Area (ΔAOB) = Area (ΔBOC) ... (equation 1)

Also, In ΔBCD, CO is the median. Therefore, median CO will divide ΔBCD into two equal triangles.

Hence, Area (ΔBOC) = Area (ΔCOD) ... (equation 2)

Similarly, Area (ΔCOD) = Area (ΔAOD) ... (equation 3)

From Equations equation (1), (2) and (3) we obtain

Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD)

Therefore, we can say that the diagonals of a parallelogram divide it into four triangles of equal area.

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 9

**Video Solution:**

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

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.3 Question 3

**Summary:**

The diagonals of a parallelogram divide it into four triangles of equal area.

**☛ Related Questions:**

- In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).
- D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that: i) BDEF is a parallelogram. ii) ar (DEF) = 1/4 ar (ABC) iii) ar (BDEF) = 1/2 ar (ABC)
- In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:i) ar (DOC) = ar (AOB)ii) ar (DCB) = ar (ACB)iii) DA || CB or ABCD is a parallelogram.[Hint: From D and B, draw perpendiculars to AC.]
- D and E are points on sides AB and AC respectively of ∆ ABC such that ar (DBC) = ar (EBC). Prove that DE || BC

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