[color=#999999][color=#999999][color=#999999]This activity belongs to the [i]GeoGebra book[/i] [url=https://www.geogebra.org/m/h3gbmymu]Linkages[/url].[/color][/color][/color][br][br]If we want to prevent degenerate cases from appearing, we should add more restrictions to the system, if possible. For example, in the previous [url=https://www.geogebra.org/m/h3gbmymu#material/sj6npzsa]four-bar construction[/url], if we want to avoid cases E' and F', where one pair of bars overlaps the other pair, we can add the black bar that appears in the following construction.[br][br]Geometrically, this bar forces the EF bar to remain horizontal at all times. We can check this fact algebraically. Taking O=(0, 0) and U=(1, 0), the above four-bar system is given by the equations:[br][list][*]E[sub]x[/sub][sup]2[/sup] + E[sub]y[/sub][sup]2[/sup] = 1[/*][*](F[sub]x[/sub] - 1)[sup]2[/sup] + F[sub]y[/sub][sup]2[/sup] = 1[/*][*](F[sub]x[/sub] - E[sub]x[/sub])[sup]2[/sup] + (F[sub]y[/sub] - E[sub]y[/sub])[sup]2[/sup] = 1[/*][/list]If we add to these equations the one corresponding to the black bar:[br][list][*]((E[sub]x[/sub]+F[sub]x[/sub])/2-1/2)[sup]2[/sup] + ((E[sub]y[/sub]+F[sub]y[/sub])/2)[sup]2[/sup] = 1[br][/*][/list]then, a simple simplification leads us to the equalities:[br][list][*]F[sub]x[/sub] - E[sub]x[/sub] = 1[/*][*]F[sub]y[/sub] - E[sub]y[/sub] = 0[/*][/list]which represent the horizontality of the bar EF, something that cannot be deduced from the previous three equations alone.

[color=#999999]Author of the construction of GeoGebra: [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color]