$ \mathrm{ABCD} $ is a square with $ \angle \mathrm{ABD}=45^{\circ} $. Find the
measures of the angles $ x, y $ and $ z $
"
Given: $ABCD$ is a square with $\angle ABD=45^{\circ}$.
To do: To find $x,\ y$ and $z$.
Solution:
As given, $ABCD$ is a square. In a square diagonals intersects each other at $90^o$.
$\Rightarrow y=\angle AOB=90^o$
In $\vartriangle OAB$,
$\angle OAB=180^o-( 45^o+90^o)$
$\angle OAB=45^o$
$AB||CD$,
$\therefore \angle CAB=\angle DCO=z=45^o$
And $x+\angle OAB=90^o$
$\Rightarrow x+45^o=90^o$
$\Rightarrow x=90^o-45^o$
$\Rightarrow x=45^o$
Thus, $x=45^o,\ y=45^o$ and $z=45^o$.
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