# Answer the 22nd question."DON'T SEND THE SIMILAR QUERY,KINDLY PERFORM THE CORRECT AND STEP BY STEP CALCULATION".

Please find below the solution to the asked query:

22 ( a ) Given : ABCD is a parallelogram , So

AB | | CD , BC | | AD and AB = CD , BC = AD --- ( 1 )

And ABEF is a parallelogram , So

AB | | EF , BE | | AF and AB = EF , BE = AF --- ( 2 )

i ) From equation 1 and 2 we take : AB | | CD , AB = CD and AB | | EF , AB = EF , So we can say :

CD | | EF ( As AB | | CD and AB | | EF )

And

CD = EF ( As AB = CD and AB = EF )

We know if a pair of sides of any quadrilateral is parallel and equal to each other then that is a parallelogram .

And

**from above two equations we can say that CDFE is a parallelogram . ( Hence proved )**

ii ) As we show CDFE is a parallelogram in previous part and we know in parallelogram opposite sides are equal and parallel to each other , So

**FD = EC --- ( 3 ) ( Hence proved )**

iii ) In $\u2206$ AFD and $\u2206$ BEC

AD = BC ( From equation 1 )

AF = BE ( From equation 2 )

And

FD = EC ( From equation 3 )

So,

**$\u2206$ AFD $\cong $ $\u2206$ BEC ( By SSS rule ) ( Hence proved )**

22 ( b ) We have our diagram , As :

Here we have join FG and AC

Given : ABCD is a parallelogram So , AB | | CD , BC | | AD and AB = CD , BC = AD --- ( 1 )

And ADEF is a square so , AD = DE = EF = AF and $\angle $ FAD = $\angle $ ADE = $\angle $ DEF = $\angle $ EFA = 90$\xb0$ --- ( 2 )

And AGHB is a square so , AG = GH = HB = AB and $\angle $ AGH = $\angle $ GHB = $\angle $ HBA = $\angle $ BAG = 90$\xb0$ --- ( 3 )

And

$\angle $ FAG = 360$\xb0$ - $\angle $ FAD - $\angle $ BAG - $\angle $ BAD , Substitute values from equation 2 and 3 we get

$\angle $ FAG = 360$\xb0$ - 90$\xb0$ - 90$\xb0$ - $\angle $ BAD ,

$\angle $ FAG = 180$\xb0$ - $\angle $ BAD

We know ABCD is a parallelogram and we know in parallelogram adjacent angles are supplementary .

$\angle $ FAG = $\angle $ ABC --- ( 4 )

In $\u2206$ FAG and $\u2206$ ABC

AF = BC ( From equation 1 and equation 2 : BC = AD and AD = DE = EF = AF )

AG = AB ( From equation 3 : AG = GH = HB = AB )

And

$\angle $ FAG = $\angle $ ABC ( From equation 4 )

So,

**$\u2206$ FAG $\cong $ $\u2206$ ABC ( By SAS rule )**

Then,

**FG = AC ( By CPCT ) ( Hence proved )**

Hope this information will clear your doubts about topic.

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