prove that two parallelograms on the same base and between same parallels are equal in area?
Answers
Two parallelograms ABCD and EFCD, on
the same base DC and between the same parallels
AF and DC are given
【TO PROVE THAT】 :-
We need to prove that ar (ABCD) = ar (EFCD).
【PROOF】 :-
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, ar (ABCD) = ar (ADE) + ar (EDCB)
= ar (BCF) + ar (EDCB) [From(5)]
= ar (EFCD)
So, parallelograms ABCD and EFCD are equal in area.
【 GIVEN】 :-
Two parallelograms ABCD and EFCD, on
the same base DC and between the same parallels
AF and DC are given
【TO PROVE THAT】 :-
We need to prove that ar (ABCD) = ar (EFCD).
【PROOF】 :-
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, ar (ABCD) = ar (ADE) + ar (EDCB)
= ar (BCF) + ar (EDCB) [From(5)]
= ar (EFCD)
So, parallelograms ABCD and EFCD are equal in area.
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