$ \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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Updated on: 10-Oct-2022

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