Please solve question no.22.b.(please solve don't send me any links....)

Dear Student,

Please find below the solution to the asked query:



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$°$                --- ( 2 )

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

And

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

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

$\angle$  FAG = 180$°$ - $\angle$ BAD

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

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

In $∆$ FAG and $∆$ 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,

$∆$ FAG $\cong$ $∆$ ABC                                             ( By SAS rule )   

Then,

FG  =  AC                                                           ( By CPCT )                              ( Hence proved )


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