When the bisectors of each angle of a quadrilateral intersect to form a second quadrilateral,

$(1)$ the sum of the two opposing angles of the quadrilateral is two right angles;

$(2)$ if the original quadrilateral is a parallelogram, the second quadrilateral is a rectangle, and its two diagonals are parallel to each side of the original quadrilateral and equal to the difference between the adjacent sides;

$(3)$ if the original quadrilateral is a rectangle, the second quadrilateral is a square.

From the diagram,

$$∠AGB=2∠R-\frac{1}{2} (∠A+∠B).$$

For the second quadrilateral $EFGH$, since $∠G=∠AGB$,

$$∠G=2∠R-\frac{1}{2} (∠A+∠B). \qquad [1]$$

Similarly,

$$∠CED=2∠R-\frac{1}{2} (∠C+∠D),$$

$$∴ \ ∠E=2∠R-\frac{1}{2} (∠C+∠D). \qquad [2]$$

From $[1]$ and $[2]$,

$$∠E+∠G=4∠R-\frac{1}{2} (∠A+∠B+∠C+∠D).$$

However, since $∠A+∠B+∠C+∠D=4∠R$,

$$∠E+∠G=4∠R-2∠R=2∠R.$$

In other words, the sum of opposite angles $∠E$ and $∠G$ is two right angles.

Therefore, the sum of another opposite angle $∠F$ and $∠H$ is also two right angles.

 

$(2)$

From the diagram,

$$∠G=2∠R-\frac{1}{2} (∠A+∠D).$$

However, since $∠A+∠D=2∠R$,

$$∠G=2∠R-∠R=∠R.$$

Similarly,

$$∠E=∠R.$$

Also,

$$∠AFB=2∠R-\frac{1}{2} (∠A+∠B).$$

However, since $∠A+∠B=2∠R$,

$$∠AFB=2∠R-∠R=∠R.$$

Furthermore, since $∠F=∠AFB$,

$$∠F=∠R.$$

Similarly,

$$∠H=∠R.$$

Therefore, since $∠E=∠F=∠G=∠H=∠R$, the quadrilateral $EFGH$ is a rectangle.

Also, since $△CDH≡△KDH$,

$$CD=KD.$$

$$AK=AD-KD,$$

$$AK=|AD-CD|.$$

However, since $FH=AK$,

$$FH=|AD-CD|.$$

In other words, the length of the diagonal of the second quadrilateral, which is a rectangle, is equal to the difference between the two adjacent sides of the original quadrilateral.

 

$(3)$