Question

prove that diagonals on a parallelogram divides it into four triangles of equal area​

Answer 1

$\huge\underline\mathfrak{Question-}$

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

$\huge\underline\mathfrak{Solution-}$

Given :

• ABCD is a parallelogram.
• AB and BD are its diagonals.

To prove :

• ar∆AEB = ar∆BEC = ar∆CED = ar∆AED

proof :

In ∆AEB and ∆CED

$\implies$ Angle BAE = Angle ECD ( alternate interior angles are equal )

$\implies$ AB = CD ( opposite sides of parallelogram are equal )

$\implies$ Angle AEB = Angle CED ( vertically opposite angles are equal )

$\therefore$ ∆AEB is congruent to ∆CED.

$\implies$ AE = CE [ corresponding parts of congruent triangles are equal ]

$\implies$ BE = DE [ corresponding parts of congruent triangles are equal ]

Similarly, ∆AEB ≅ ∆BEC ≅ ∆CED ≅ ∆DEA

★ $\pink{Congruent\:triangles\:are\:equal\:in\:area}$

Therefore, diagonals of a parallelogram divides it into four triangles of equal area.

Hence proved!

Answer 2

Let us consider in a parallelogram ABCD the diagonals AC and BD are cut at point O.

To prove: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)

Proof:

In parallelogram ABCD the diagonals bisect each other.

AO = OC

In ∆ACD, O is the mid-point of AC. DO is the median.

ar (∆AOD) = ar (COD) ….. (1) [Median of ∆ divides it into two triangles of equal arreas]

Similarly, in ∆ ABC

ar (∆AOB) = ar (∆COB) ….. (2)

In ∆ADB

ar (∆AOD) = ar (∆AOB) …. (3)

In ∆CDB

ar (∆COD) = ar (∆COB) …. (4)

From (1), (2), (3) and (4)

ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)

Hence proved.

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