# A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)

**Solution:**

Given, ABCD is a parallelogram

A point E is taken on the side BC

AE and DC are produced to meet at F

We have to prove that ar(ADF) = ar(ABFC)

Join the diagonal AC

Considering triangles ABE and CFE,

Since E is the midpoint of BC

BE = CE

We know that the vertically opposite angles are equal

So, ∠AEB = ∠CEF

Since AB || CD and CD is produced to CF

∠ABE = ∠FCE

ASA criterion states that two triangles are congruent if any two angles and the side included between them of one triangle are equal to the corresponding angles and the included side of the other triangle.

By ASA criterion, the triangles ABE and CFE are congruent.

So, AB = CF

Given, ABCD is a parallelogram.

We know that the opposite sides of a parallelogram are parallel and congruent.

So, AB = CD

Therefore, AB = CF = CD ------------ (1)

Considering parallelogram ABFC,

Since C is the midpoint of FD

AF is the diagonal of ABFC

So, ar(ABF) = ar(ACF) --------------- (2)

In triangle ADF,

From (1), CD = CF

AC is the median.

We know that the median of a triangle divides it into two triangles of equal areas.

So, ar(ACF) = ar(ACD) --------------- (3)

From (2) and (3)

ar(ABF) = ar(ACD)

Adding ar(ACF) on both sides,

ar(ABF) + ar(ACF) = ar(ACD) + ar(ACF)

From the figure,

ar(ABF) + ar(ACF) = ar(ABCF)

So, ar(ABCF) = ar(ACD) + ar(ACF)

From the figure,

ar(ACD) + ar(ACF) = ar(ADF)

Therefore, ar(ABCF) = ar(ADF)

**✦ Try This: **In the adjoining figure, BD || CA, E is the midpoint of CA and BD = 1/2 CA. Prove that ar (△ ABC) = 2 ar (△ DBC).

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

**NCERT Exemplar Class 9 Maths Exercise 9.4 Problem 1**

## A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)

**Summary:**

A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. It is proven that ar (ADF) = ar (ABFC)

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