In $ \triangle \mathrm{ABC} $ and $ \triangle \mathrm{DEF}, \mathrm{AB}=\mathrm{DE}, \mathrm{AB} \| \mathrm{DE}, \mathrm{BC}=\mathrm{EF} $ and $ \mathrm{BC} \| EF $. Vertices $ \mathrm{A}, \mathrm{B} $ and $ \mathrm{C} $ are joined to vertices D, E and F respectively (see below figure).

Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral $ \mathrm{BEFC} $ is a parallelogram
(iii) $ \mathrm{AD} \| \mathrm{CF} $ and $ \mathrm{AD}=\mathrm{CF} $
(iv) quadrilateral ACFD is a parallelogram
(v) $ \mathrm{AC}=\mathrm{DF} $
(vi) $ \triangle \mathrm{ABC} \equiv \triangle \mathrm{DEF} $."

AcademicMathematicsNCERTClass 9

Given:

In \( \triangle \mathrm{ABC} \) and \( \triangle \mathrm{DEF}, \mathrm{AB}=\mathrm{DE}, \mathrm{AB} \| \mathrm{DE}, \mathrm{BC}=\mathrm{EF} \) and \( \mathrm{BC} \| EF \).

To do :

We have to show that

(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral \( \mathrm{BEFC} \) is a parallelogram
(iii) \( \mathrm{AD} \| \mathrm{CF} \) and \( \mathrm{AD}=\mathrm{CF} \)
(iv) quadrilateral ACFD is a parallelogram
(v) \( \mathrm{AC}=\mathrm{DF} \)
(vi) \( \triangle \mathrm{ABC} \equiv \triangle \mathrm{DEF} \).

Solution :  

(i) In quadrilateral $ABED$,

$AB=DE$

$AB \| DE$

We know that,

A parallelogram is a quadrilateral with opposite sides equal and parallel to each other.

Therefore,

$ABED$ is a parallelogram.

(ii) In quadrilateral $BECF$,

$BC=EF$

$BC \| EF$

Therefore,

$BECF$ is a parallelogram.

(iii) $ABED$ and $BEFC$ are parallelograms.

We know that,

Opposite sides of a parallelogram are equal and parallel.

This implies,

$AD = BE$

$BE = CF$

$\Rightarrow AD = CF$

$AD \| BE$

$BE \| CF$

$\Rightarrow AD \| CF$

(iv) In quadrilateral $ACFD$,

$AD=CF$         (Proved)

$AD \| CF$         (Proved)

Therefore,

$ACFD$ is a parallelogram.

(v) $ACFD$ is a parallelogram

This implies,

$AC \| DF$

$AC = DF$

(vi) In $\triangle ABC$ and $\triangle DEF$,

$AB = DE$      (Given)

$BC = EF$        (Given)

$AC = DF$        (Opposite sides of a parallelogram are equal)

Therefore, by SSS congruency, we get,

$\triangle ABC \cong \triangle DEF$.

raja
Updated on 10-Oct-2022 13:41:02

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