Peaucellier–Lipkin linkage

First, it must be proven that points O, B, D are collinear. This may be easily seen by observing that the linkage is mirror-symmetric about line OD, so point B must fall on that line.

More formally, triangles △BAD and △BCD are congruent because side BD is congruent to itself, side BA is congruent to side BC , and side AD is congruent to side CD . Therefore, angles ∠ABD and ∠CBD are equal.

Next, triangles △OBA and △OBC are congruent, since sides OA and OC are congruent, side OB is congruent to itself, and sides BA and BC are congruent. Therefore, angles ∠OBA and ∠OBC are equal.

Finally, because they form a complete circle, we have

${\displaystyle \angle OBA+\angle ABD+\angle DBC+\angle CBO=360^{\circ }}$

but, due to the congruences, ∠OBA = ∠OBC and ∠DBA = ∠DBC, thus

${\displaystyle {\begin{aligned}&2\times \angle OBA+2\times \angle DBA=360^{\circ }\\&\angle OBA+\angle DBA=180^{\circ }\end{aligned}}}$

therefore points O, B, and D are collinear.

Let point P be the intersection of lines AC and BD. Then, since ABCD is a rhombus, P is the midpoint of both line segments BD and AC. Therefore, length BP = length PD.

Triangle △BPA is congruent to triangle △DPA, because side BP is congruent to side DP, side AP is congruent to itself, and side AB is congruent to side AD . Therefore, angle ∠BPA = angle ∠DPA. But since ∠BPA + ∠DPA = 180°, then 2 × ∠BPA = 180°, ∠BPA = 90°, and ∠DPA = 90°.

Let:

${\displaystyle {\begin{aligned}&x=\ell _{BP}=\ell _{PD}\\&y=\ell _{OB}\\&h=\ell _{AP}\end{aligned}}}$

Then:

${\displaystyle \ell _{OB}\cdot \ell _{OD}=y(y+2x)=y^{2}+2xy}$

${\displaystyle {\ell _{OA}}^{2}=(y+x)^{2}+h^{2}}$ (due to the Pythagorean theorem)

${\displaystyle {\ell _{OA}}^{2}=y^{2}+2xy+x^{2}+h^{2}}$(same expression expanded)

${\displaystyle {\ell _{AD}}^{2}=x^{2}+h^{2}}$ (Pythagorean theorem)

${\displaystyle {\ell _{OA}}^{2}-{\ell _{AD}}^{2}=y^{2}+2xy=\ell _{OB}\cdot \ell _{OD}}$

Since OA and AD are both fixed lengths, then the product of OB and OD is a constant:

${\displaystyle \ell _{OB}\cdot \ell _{OD}=k^{2}}$

and since points O, B, D are collinear, then D is the inverse of B with respect to the circle (O,k) with center O and radius k.

Thus, by the properties of inversive geometry, since the figure traced by point D is the inverse of the figure traced by point B, if B traces a circle passing through the center of inversion O, then D is constrained to trace a straight line. But if B traces a straight line not passing through O, then D must trace an arc of a circle passing through O. Q.E.D.

Slider-rocker four-bar acts as the driver of the Peaucellier–Lipkin linkage

Peaucellier–Lipkin linkages (PLLs) may have several inversions. A typical example is shown in the opposite figure, in which a rocker-slider four-bar serves as the input driver. To be precise, the slider acts as the input, which in turn drives the right grounded link of the PLL, thus driving the entire PLL.

Sylvester (Collected Works, Vol. 3, Paper 2) writes that when he showed a model to Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it—it is the most beautiful thing I have ever seen in my life.’”