The statement “Every Diagonal of a Parallelogram Divides it into Two Triangles of Equal Area” is a property related to the area of the triangle and area of a parallelogram. By using the Every Diagonal of a Parallelogram Divides it into Two Triangles of Equal Area theorem the students of 9th grade can solve different types of problems related to the concept Area.

Also Read:

- Base and Height (Altitude) in a Triangle and a Parallelogram
- If Each Diagonal of a Quadrilateral Divides it in Two Triangles of Equal Area then Prove that the Quadrilateral is a Parallelogram

## Every Diagonal of a Parallelogram Divides it into Two Triangles of Equal Area – Theorem

Statement:

Prove that Every Diagonal of a Parallelogram Divides it into Two Triangles of Equal Area.

Proof:

Here ABCD is a parallelogram and BD is diagonal

In ∆ADB and ∆CBD

AD // BC

∠ADB = ∠CBD( alternate angles )

Also AB // DC

∠ABD = ∠CDB( alternate angles )

DB = BD

Therefore ∆ADB ≅ ∆CDB( by SAS congruence rule)

Since congruent figures have the same area

ar(ADB) = ar(CBD)

Hence proved.

### FAQs on A Diagonal Divides a Parallelogram into Two Triangles of Equal Area

**1. What divides a parallelogram into two triangles of equal area?**

The diagonal divides a parallelogram into two triangles of equal area.

**2. Are the opposite sides of a parallelogram always equal?**

opposite sides of a parallelogram are equal. The diagonals of a parallelogram bisect each other. Each diagonal of a parallelogram bisects into two congruent triangles.The quadrilateral is a parallelogram.

**3. Do all parallelograms have 4 equal sides?**

A parallelogram has two parallel pairs of opposite sides. A rectangle has two pairs of opposite sides parallel, and it has four right angles. It is also a parallelogram and it has two pairs of parallel sides. A square has two pairs of parallel sides, four right angles, and all four sides are equal.