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In $\Delta A B C $ and $\Delta D E F$ , A B=D E, $A B \parallel D E$, $B C=E F$ and $BC \parallel EF$. Vertices A, B and C are joined to vertices D, E and F respectively.
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) $A D \parallel C F$ and AD=CF
(iv) quadrilateral ACFD is a parallelogram
(v) $AC=DF$
(vi) $\Delta ABC \cong \Delta DEF$"
Given :
$AB = DE$ , $AB || DE$, $BC \parallel EF$.
To do :
We have to show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) $A D \parallel C F$ and AD=CF
(iv) quadrilateral ACFD is a parallelogram
(v) $AC=DF$
(vi) $\Delta ABC \cong \Delta DEF$
Solution :
(i) AB = DE.
If a pair of opposite sides is equal and parallel, then the quadrilateral is a parallelogram.
Therefore, ABED is a parallelogram.
(ii) BC = EF and BC || EF.
If a pair of opposite sides is equal and parallel, then the quadrilateral is a parallelogram.
Therefore, BEFC is a parallelogram.
(iii) In parallelogram ABED,
AD = BE and AD || BE.
In parallelogram BECF,
BE = CF and BE || CF.
∴ AD = BE and BE = CF, then AD = CF.
AD || BE and BE || CF,
Therefore, AD || CF.
(iv)AD = CF and AD || CF.
Therefore, ACFD is a parallelogram.
(v) AC and DF are opposite sides of the parallelogram ACFD.
∴ AC=DF.
(vi) AB = DE and BC = EF (given)
AC = DF
∴∆ABC ≅ ∆DEF (By S.S.S congruency)
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