# Please solve this question Dear Student,

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 $∆$ AFD and $∆$ BEC

AD =  BC                                                           ( From equation 1 )

AF =  BE                                                            ( From equation 2 )

And

FD =  EC                                                            ( From equation 3 )

So,

$∆$ AFD $\cong$ $∆$ 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$°$                --- ( 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 )