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Parallelograms on the same base and between the same parallels have the same area

Theorem - 1
Parallelograms on the same base and between the same parallels are equal in area.
Given: Two parallelogram ABCD and EFCD are on the same base DC and between the same parallel lines AF and DC.
To Prove: ar(||gm ABCD) = ar( ||gm EFCD)

 

Parallelograms on the same base and between the same parallels are equal in area

Proof :

In ΔAED and ΔBFC, we have
ED = FC ...(opposite sides of ||gm)
AD = BC ...(opposite sides of ||gm)

EDA = FCB ...(AD || BC and ED || FC)
ΔAED ΔBFC (SAS Congruence Criteria)
ar(
ΔAED) = ar (ΔBFC ) (By Congruence area axiom)
ar(||gm ABCD) = ar(quadrilateral EBCD) + ar(
ΔAED) ...(i) (area addition axiom)
ar(||gm EFCD) = ar(quadrilateral. EBCD) + ar(
ΔBFC) ...(ii) (area addition axiom)
ar(||gm ABCD) = ar(||gm EFCD)

 
Example :

Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram, divides it into two parallelograms of equal area.

Solution

In ||gm ABCD, E is the mid-point of AB and F is the mid-point of DC. 
Also AB|| DC.

 Line segment joining the mid-points of a pair of opposite sides of a parallelogram,
divides it into two parallelograms of equal area.


AB = DC and AB || DC
 
AB = DC and AE || DF (Since E and F mid point of AB and DC)

∴ AE =

AB and DF = DC
AE = DF and AE || DF
Quadrilateral AEFD is a parallelogram

Similarly, € EBCF is a parallelogram.

Now parallelogram AEFD and EBCF are on equal bases DF = FC and between two parallels AB and DC

ar(||gm AEFD) = ar(||gm EBCF)
 




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