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Parallelogram on the same base and between the same parallels are equal in area. Prove!

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Answer :

nadafathima
Theorem : Parallelograms on the same base and between the same parallels are equal in area.
Given:  Two parallelograms ABCD and EFCD
To prove: ar (ABCD) = ar (EFCD).
Proof : Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels.
In Δ ADE and Δ BCF,
∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)
∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2) Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)
Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)
So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]
Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)
Now,adding ar (EDCB) both the sides,
 ar (ADE) + ar (EDCB) = ar (BCF)+ ar (EDCB) ar (ABCD) = ar (EFCD)
 So, parallelograms ABCD and EFCD are equal in area.
nalavathramchand

Answer:

Step-by-step explanation:Theorem : Parallelograms on the same base and between the same parallels are equal in area.

Given:  Two parallelograms ABCD and EFCD

To prove: ar (ABCD) = ar (EFCD).

Proof : Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels.

In Δ ADE and Δ BCF,

∠ DAE = ∠ CBF (Corresponding angles from AD || BC and transversal AF) (1)

∠ AED = ∠ BFC (Corresponding angles from ED || FC and transversal AF) (2) Therefore, ∠ ADE = ∠ BCF (Angle sum property of a triangle) (3)

Also, AD = BC (Opposite sides of the parallelogram ABCD) (4)

So, Δ ADE ≅ Δ BCF [By ASA rule, using (1), (3), and (4)]

Therefore, ar (ADE) = ar (BCF) (Congruent figures have equal areas) (5)

Now,adding ar (EDCB) both the sides,

 ar (ADE) + ar (EDCB) = ar (BCF)+ ar (EDCB) ar (ABCD) = ar (EFCD)

 So, parallelograms ABCD and EFCD are equal in area.

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