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

Answer the 22nd question."DON'T SEND THE SIMILAR QUERY,KINDLY PERFORM THE CORRECT AND STEP BY STEP CALCULATION". But z 1 + = 909 z 1 + z2 = 900 22. (a) In the figure (1) given below, ABCD and ABEF are parallelograms. Prove that (i) CDFE is a parallelogram (ii) FD = EC (iii) AAFD ABEC. (b) In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG AC. F (2) (1) ABII DC, AB II FEZ* DC II FE,AB = DC, 23. (b) AAFG= ABCA, for, AF = BC,AG =AB, ZFAG = 3600 — 900 - 900 — z A = 1800 — = LB. mbus in which ZA = 600. Find the ratio AC : BD. O

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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 $\cong$ $∆$ 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 )


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