# 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 Î”AED and Î”BFC, we have

ED = FC ...(opposite sides of ||gm)

AD = BC ...(opposite sides of ||gm)

âˆ EDA = âˆ FCB ...(AD || BC and ED || FC)

Î”AED â‰… Î”BFC (SAS Congruence Criteria)

ar(Î”AED) = ar (Î”BFC ) (By Congruence area axiom)

ar(||gm ABCD) = ar(quadrilateral EBCD) + ar(Î”AED) ...(i) (area addition axiom)

ar(||gm EFCD) = ar(quadrilateral. EBCD) + ar(Î”BFC) ...(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, |

AB = DC and AB || DC

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

âˆ´ AE =

AB and DF = DCâˆ´ AE = DF and AE || DF

âˆ´Quadrilateral AEFD is a parallelogram

Similarly, â‚¬ EBCF is a parallelogram.

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

âˆ´ ar(||gm AEFD) = ar(||gm EBCF)