Question

a. The bisector of the angles of $▱\text{ABCD}$ intersect to form quad. $\text{WXYZ}$. What special kind of quadrilateral is $\text{WXYZ}$.

b. Prove the answer to part (a).

Short answer

a. Quad. $\text{WXYZ}$ is a rectangle.

b. It is proved that Quad. $\text{WXYZ}$is a rectangle.

Step-by-step solution

a. Step 1 - Consider the diagram.

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

- State the explanation.

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

- State the conclusion.

Therefore, quad. $WXYZ$ is a rectangle.

b. Step 1 - Consider the diagram.

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

- State the proof.

Angles A and D are bisected, therefore,

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

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

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

Then,

$\angle DAW+\angle WDA=90$ …… (1)

In $\Delta AWD$,

$\angle DAW+\angle WDA+\angle DWA=180$

From (1)

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

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

Similarly,

$\angle WZY=\angle ZYX=\angle YXW=90$

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

- State the conclusion.

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