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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