Question

ABCD is a quadrilateral in which AB =CD and AD=BC. Show that it is a parallelogram.

Answer 1

$<b>Heya !$

Here's your answer !!

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Given That :

In a quadrilateral ABCD -

AB = CD

AD = BC


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Construction :

Draw a diagonals AC and BD.


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Proof :

In ∆ ABC and ∆ ADC ,

AB = CD { given }

AD = BC { given }

AC = CA { common }

By S.S.S. criteria,

∆ABC is congruent to ∆ADC

$\angle{\bf{ B }}$ = $\angle{\bf{D}}$ { c.p.c.t. }


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In ∆ ABD and ∆ CDB ,

AB = CD { given }

AD = BC { given }

BD = DB { common }

By S.S.S. criteria,

∆ABD is congruent to ∆CDB

$\angle{\bf{ A }}$ = $\angle{\bf{C}}$ { c.p.c.t. }


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We know, of pair of opposite angles of a quadrilateral are equal, then it is a parallelogram.

Since ,

$\angle{\bf{ B }}$ = $\angle{\bf{D}}$

$\angle{\bf{ A }}$ = $\angle{\bf{C}}$

which are pairs of opposite angles of the quadrilateral.



Therefore,

ABCD is a parallelogram.

[ Hence Proved ]

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Thanks !!

Answer 2

Heyy Mate❤✌✌❤

given , 
AB = CD 

AD = BC

☆ to prove that : ABCD is a IIgram


☆ proof :- AB = CD (from the given )

              BC = AD (from the given )

              BD = BD (common side )

∴ Δ ADB ≡ Δ DBC by SSS congruence rule . 
therefore from CPCT we can write that ∠ADB = ∠ DBC (alternate interior angle )

since the alternate interior angles and opposite sides are equal in a quadrilateral therefore it is a IIgram .
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