Consider the following two statements:

Statement 1: The line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

Statement 2: Diagonals of a parallelogram divide it into four triangles of equal area.

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Let ABCD be a parallelogram in which E and F are mid-points of AB and CD respectively. Join EF.

Let us construct DG ⊥ AB and let DG = h, where h is the altitude on side AB.

Area of || gm ABCD = base × height = AB × h

Area of ||gm AEFD = AE × h = $\dfrac{AB}{2}$ × h ……………….(1) [Since E is the mid-point of AB]

Area of ||gm EBCF = EB × h = $\dfrac{AB}{2}$ × h ……………….(2) [Since E is the mid-point of AB]

From (1) and (2),

Area of || gm AEFD = Area of || gm EBCF.

Thus, the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.

∴ Statement 1 is true.

From figure,

The diagonals AC and BD cut at point O.

In parallelogram, the diagonals bisect each other.

∴ AO = OC

In ∆ACD, O is the mid-point of AC.

∴ OD is the median.

Area of ∆AOD = Area of ∆COD …………….. (3) [Median of ∆ divides it into two triangles of equal areas.]

Similarly, in ∆ABC

O is the mid-point of AC.

∴ OB is the median.

Area of ∆AOB = Area of ∆COB …………….. (4) [Median of ∆ divides it into two triangles of equal areas.]

In ∆ADB,

O is the mid-point of BD.

∴ OA is the median.

Area of ∆AOD = Area of ∆AOB …………….. (5)

From (3), (4) and (5) we get,

Area of ∆AOB = Area of ∆COB = Area of ∆COD = Area of ∆AOD

So proved, that the diagonals of a parallelogram divide it into four triangles of equal area.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.