91Ó°ÊÓ

Here, $ABCD$ is a parallelogram and quad. $WXYZ$ is formed by bisector of the angles of parallelogram $ABCD$.

Quad. $WXYZ$ is a rectangle because intersection angle will be ${90}^{∘}$.

Therefore, quad. $WXYZ$ is a rectangle.

Here, $ABCD$ is a parallelogram and quad. $WXYZ$ is formed by bisector of the angles of parallelogram $ABCD$.

Angles A and D are bisected, therefore,

$∠DAW=\frac{1}{2}∠A$and $∠WDA=\frac{1}{2}∠D$

Since A and D is supplementary angle of parallelogram, therefore,

$\begin{array}{c}∠A+∠D=180\\ 2∠DAW+2∠WDA=180\end{array}$

Then,

$∠DAW+∠WDA=90$ …… (1)

In $\mathrm{Δ}AWD$,

$∠DAW+∠WDA+∠DWA=180$

From (1)

$\begin{array}{c}90+∠DWA=180\\ ∠DWA=90\end{array}$

Therefore, $∠XWZ=90$ (As $∠DWA$ and $∠XWZ$are vertically opposite angles.)

Similarly,

$∠WZY=∠ZYX=∠YXW=90$

Since all the angles in $WXYZ$ are ${90}^{∘}$. Then it is a rectangle.

Therefore, quad. $WXYZ$ is a rectangle (proved).