Let $a$ and $b$ be two adjacent sides of the original parallelogram, and $2 \alpha \leq 90^{\circ}$ the angle between them. Let $x$ and $y$ be two adjacent sides of the parallelogram formed by the angle bisectors. Then, as can be seen from a sketch,

$$
x=a \sin \alpha-b \sin \alpha=(a-b) \sin \alpha \quad ; \quad y=a \cos \alpha-b \cos \alpha=(a-b) \cos \alpha
$$

For the intersection angle $\varphi$ of two non-parallel angle bisectors, the angle sum theorem for plane triangles gives

$$
\alpha+0.5\left(180^{\circ}-2 \alpha\right)+\varphi=180^{\circ} \rightarrow \varphi=90^{\circ}
$$

i.e., the resulting parallelogram is a rectangle. Thus, for its area $A_{2}$,

$$
A_{2}=x y=(a-b)^{2} \sin \alpha \cos \alpha=0.5(a-b)^{2} \sin (2 \alpha)
$$

For the area $A_{1}$ of the original parallelogram, $A_{1}=a b \sin (2 \alpha)$. Therefore,

$$
A_{2}: A_{1}=\frac{0.5(a-b)^{2} \sin (2 \alpha)}{a b \sin (2 \alpha)}=\frac{(a-b)^{2}}{a b}
$$

Remark: If the external angle bisectors are used instead of the (internal) angle bisectors, then analogously $A_{2}: A_{1}=(a+b)^{2}:(2 a b)$.
## Task 16/86

The angle bisectors of a parallelogram determine another parallelogram. Determine the ratio of the areas of this and the original parallelogram.