PQRS is a rhombus. XPQY is a straight line such that XP = PQ = QY . IF XS and YR are produced to meet at point Z, the measure of angle XYZ is  

 

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Please find below the solution to the asked query:

We form our diagram from given information , As :

Here PQ =  XP  =  QY =  QR  =  RS = SP                --- ( A ) ( As PQRS is a rhombus ) and PQ | | RS and QR | | SP 

Here , 

$\angle$ SPX =  $\angle$ PQR                  --- (1 ) ( Corresponding angle as QR | | SP and XY is transversal line )

And

$\angle$ YQR =  $\angle$ QPS                  --- ( 2 ) ( Corresponding angle as QR | | SP and XY is transversal line )

In $∆$ XPS , XP  =  PS  from equation " A"  , So from base angle theorem we get 

$\angle$ PXS =  $\angle$ PSX                  --- ( 3 ) 

And from angle sum property of triangle we get in $∆$  XPS 

$\angle$ PXS +  $\angle$ PSX  +  $\angle$ SPX =  180$°$ , Now substitute values from equation 1 and 3 we get

$\angle$ PXS  +  $\angle$ PXS  + $\angle$ PQR =  180$°$

2 $\angle$ PXS +  $\angle$PQR =  180$°$

2 $\angle$ PXS = 180$°$ - $\angle$ PQR  

2 $\angle$ PXS = $\angle$ QPS                                      --- ( 4 )      ( We know adjacent angles are supplementary in rhombus )

And

In $∆$ YQR , QY  =  QR  from equation " A"  , So from base angle theorem we get 

$\angle$ QYR =  $\angle$ QRY                                   --- ( 5 ) 

And from angle sum property of triangle we get in $∆$  YQR

$\angle$ QYR +  $\angle$ QRY +  $\angle$ YQR =  180$°$ , Now substitute values from equation 2 and 5 we get

$\angle$ QYR +  $\angle$ QYR + $\angle$ QPS =  180$°$

2 $\angle$ QYR  + 2 $\angle$ PXS =  180$°$  , From equation 4

2 ( $\angle$ QYR + $\angle$ PXS ) = 180$°$

$\angle$ QYR + $\angle$ PXS = 90$°$

$\angle$ XYZ + $\angle$ YXZ = 90$°$                                      --- ( 6 )    (  We know $\angle$ QYR =  $\angle$ XYZ and $\angle$ PXS = $\angle$ YXZ same angles )

Now from angle sum property of triangle we get in $∆$  XYZ

$\angle$ XYZ +  $\angle$ YXZ +  $\angle$ XZY =  180$°$ , Now substitute values from equation 6 we get

90$°$ + $\angle$ XZY =  180$°$

$\angle$ XZY =  90$°$                                                                  (  Ans )
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