[tex]\large\underline{\sf{Solution-}}[/tex]
Let assume that ABCD and EFCD be two parallelogram on the same base CD and between same parallels.
As, AD || BC and AB is transversal.
We know, Corresponding angles are equal.
[tex]\implies\sf\:\angle DAE = \angle CBF \\ [/tex]
Also, CF || DE and AB is transversal.
[tex]\implies\sf\:\angle DEA = \angle CFB \\ [/tex]
Now, In triangle DAE and triangle CBF
[tex]\boxed{\begin{aligned}& \:\sf \: \angle DAE = \angle CBF \: \{proved \: above \} \\ \\& \:\sf \: \angle DAE = \angle CBF \: \{proved \: above \} \\ \\& \:\sf \: AD=BC \: \{Opposite \: sides \}\end{aligned}} \implies\sf\:\triangle DAE \:\cong \:\triangle CBF \\ [/tex]
[ By AAS Congruency
We know, if two triangles are congruent, then they have same area.
[tex] \implies\sf\:Area \: (\triangle DAE) \: = \:Area \: (\triangle CBF) \\ [/tex]
On adding area of quadrilateral DEBC on both sides, we get
[tex] \sf\:Area \: (\triangle DAE) + Area(DEBC) \: = \:Area \: (\triangle CBF) + Area(DEBC) \\ [/tex]
[tex]\implies\sf\:Area \: ( { \parallel}^{gm}\:ABCD) = Area \: ( { \parallel}^{gm}\: EFCD) \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:Area \: ( { \parallel}^{gm}\:ABCD) = Area \: ( { \parallel}^{gm}\: EFCD) \: } \\ [/tex]
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[tex]\large\underline{\sf{Solution-}}[/tex]
Let assume that ABCD and EFCD be two parallelogram on the same base CD and between same parallels.
As, AD || BC and AB is transversal.
We know, Corresponding angles are equal.
[tex]\implies\sf\:\angle DAE = \angle CBF \\ [/tex]
Also, CF || DE and AB is transversal.
We know, Corresponding angles are equal.
[tex]\implies\sf\:\angle DEA = \angle CFB \\ [/tex]
Now, In triangle DAE and triangle CBF
[tex]\boxed{\begin{aligned}& \:\sf \: \angle DAE = \angle CBF \: \{proved \: above \} \\ \\& \:\sf \: \angle DAE = \angle CBF \: \{proved \: above \} \\ \\& \:\sf \: AD=BC \: \{Opposite \: sides \}\end{aligned}} \implies\sf\:\triangle DAE \:\cong \:\triangle CBF \\ [/tex]
[ By AAS Congruency
We know, if two triangles are congruent, then they have same area.
[tex] \implies\sf\:Area \: (\triangle DAE) \: = \:Area \: (\triangle CBF) \\ [/tex]
On adding area of quadrilateral DEBC on both sides, we get
[tex] \sf\:Area \: (\triangle DAE) + Area(DEBC) \: = \:Area \: (\triangle CBF) + Area(DEBC) \\ [/tex]
[tex]\implies\sf\:Area \: ( { \parallel}^{gm}\:ABCD) = Area \: ( { \parallel}^{gm}\: EFCD) \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:Area \: ( { \parallel}^{gm}\:ABCD) = Area \: ( { \parallel}^{gm}\: EFCD) \: } \\ [/tex]