1. **Problem statement:** We have a parallelogram with sides $a$ and $b$ and an angle $\alpha$ between them. Internal bisectors of the angle are drawn, and we need to find the area of the resulting rectangle formed by these bisectors. 2. **Understanding the problem:** The internal bisectors of the angles of a parallelogram intersect to form a rectangle. We want to find the area of this rectangle. 3. **Key formulas and facts:** - Area of parallelogram: $S = ab \sin \alpha$ - The internal bisectors of adjacent angles in a parallelogram form a rectangle whose sides are the bisectors of the angles. 4. **Finding the lengths of the bisectors:** The bisector of an angle $\theta$ in a triangle with adjacent sides $x$ and $y$ has length given by: $$l = \frac{2xy \cos(\frac{\theta}{2})}{x + y}$$ In our parallelogram, the bisectors correspond to angles $\alpha$ and $\pi - \alpha$. 5. **Calculate the bisector lengths:** - For angle $\alpha$, bisector length: $$l_1 = \frac{2ab \cos(\frac{\alpha}{2})}{a + b}$$ - For angle $\pi - \alpha$, bisector length: $$l_2 = \frac{2ab \cos(\frac{\pi - \alpha}{2})}{a + b} = \frac{2ab \sin(\frac{\alpha}{2})}{a + b}$$ 6. **Area of the rectangle formed by bisectors:** $$A = l_1 \times l_2 = \left(\frac{2ab \cos(\frac{\alpha}{2})}{a + b}\right) \times \left(\frac{2ab \sin(\frac{\alpha}{2})}{a + b}\right) = \frac{4a^2 b^2 \cos(\frac{\alpha}{2}) \sin(\frac{\alpha}{2})}{(a + b)^2}$$ 7. **Simplify using double angle identity:** $$\sin \alpha = 2 \sin(\frac{\alpha}{2}) \cos(\frac{\alpha}{2})$$ So, $$A = \frac{4a^2 b^2}{(a + b)^2} \times \frac{\sin \alpha}{2} = \frac{2a^2 b^2 \sin \alpha}{(a + b)^2}$$ **Final answer:** $$\boxed{A = \frac{2a^2 b^2 \sin \alpha}{(a + b)^2}}$$ This is the area of the rectangle formed by the internal bisectors of the parallelogram.