Show that parallelograms on the same base between same parallel lines are
equal in area.
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Answer:
Theorem: Parallelograms on the same base and between the same parallels are equal in area.
Proof: Consider the figure presented above. Can you see that
Δ
B
C
E
ΔBCE
and
Δ
A
D
F
ΔADF
will be congruent? This is easy to show. We have:
BC = AD (opposite sides of a parallelogram are equal)
∠
B
C
E
∠ B C E
=
∠
A
D
F
∠ A D F
(corresponding angles)
∠
B
E
C
∠ B E C
=
∠
A
F
D
∠ A F D
(corresponding angles)
By the ASA criterion, the two triangles are congruent, which means that their areas are equal. Now,
area(ABCD) = area(ABED) + area(
Δ
B
C
E
ΔBCE
)
Similarly,
area(ABEF) = area(ABED) + area(
Δ
A
D
F
ΔADF
)
Clearly,
area(ABCD) = area(ABEF)
This completes the proof.
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