Please solve this question

Please solve this question S. A BCD is a parallelogram. ADEF and AGHB are two squares. Prove that FG AC.

Dear Student,

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 AFD and BEC

AD =  BC                                                           ( From equation 1 )

AF =  BE                                                            ( From equation 2 )

And

FD =  EC                                                            ( From equation 3 )

So,

AFD 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  FAD =   ADE =  DEF =  EFA = 90°                --- ( 2 )

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

And

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

  FAG = 360° - 90° - 90° BAD ,

  FAG = 180° BAD

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

  FAG = 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

  FAG = ABC                                             ( From equation 4 )

So,

FAG ABC                                             ( By SAS rule )   

Then,

FG  =  AC                                                           ( By CPCT )                              ( Hence proved )


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