Straight Line Mechanism due to Hart (1874)

Points A and B are fixed

Links AD and BC are pivoted at A and B, such that AD = BC = AB = a.

Links DP, CP and ST are pivoted at their ends and are equal in length = b.

Also, DS.DA = DP ² ( and CT.CB = CP ² )   [CT = DS = c]

Show that P lines on the perpedicular bisector of AB.

A complex approach (A-level)

Let the angle BAD be called α.
Let the angle ABC be called π − β.
Let the angle between DP and the horizontal (i.e. parallel to AB) be called θ.
Let the angle between CP and the horizontal be called φ.

Let us map this onto the Argand diagram with its origin O at the mid−point of AB and with the x-axis along OB.
This choice of angles makes them all positive in the sense of the Argand diagram.

      A = − ½ a
      D = A + a eⁱα = − ½ a + a eⁱα
      S = − ½ a + (a − c) eⁱα
      B = + ½ a
      C = ½ a + a eⁱβ
      T = ½ a + (a − c) eⁱβ

      P = D + b eⁱθ = − ½ a + a eⁱα + b eⁱθ
      P = ½ a + a eⁱβ + b eⁱφ

The rod ST of length b forces there to be a functional relationship between α and β.
      T − S = a + (a − c) (eⁱβ − eⁱα)
      a c = b ² = (T − S)(T − S)*

A bit of slog algebra ...
      a c = [a + (a − c) (eⁱβ − eⁱα)][a + (a − c) (e^-iβ − e^-iα)]
            = a ² + (a − c) ² (eⁱβ − eⁱα) (e^-iβ − e^-iα)   +   a (a − c) (eⁱβ − eⁱα + e^-iβ − e^-iα)
            = a ² + (a − c) ² (1 − eⁱ(α-β) − e^-i(α-β) + 1)   +   a (a − c) (eⁱβ + e^-iβ − eⁱα − e^-iα)
            = a ² + 2 (a − c) ² − (a − c) ² (eⁱ(α-β) + e^-i(α-β))   +   a (a − c) (eⁱβ + e^-iβ − eⁱα − e^-iα)
       0   = 3 a ² − 5 a c + 2 c ² − (a − c) ² (eⁱ(α-β) + e^-i(α-β))   +   a (a − c) (eⁱβ + e^-iβ − eⁱα − e^-iα)
       0   = (a − c) (3 a − 2 c) − (a − c) ² (eⁱ(α-β) + e^-i(α-β))   +   a (a − c) (eⁱβ + e^-iβ − eⁱα − e^-iα)
       0   = 3 a − 2 c − (a − c) (eⁱ(α-β) + e^-i(α-β))   +   a (eⁱβ + e^-iβ − eⁱα − e^-iα)
       0   = a + (a − c) (2 − eⁱ(α-β) − e^-i(α-β))   +   a (eⁱβ + e^-iβ − eⁱα − e^-iα)
... leads to:
       0   = a + 2 (a − c) (1 − cos(α − β))   +   2 a (cos β − cos α)

Now switch our attention to the two alternative expressions for P:
      P − P = 0 = a + a (eⁱβ − eⁱα) + b (eⁱφ − eⁱθ)
      a + a (eⁱβ − eⁱα) = − b (eⁱφ − eⁱθ)
      [a + a (eⁱβ − eⁱα)] [a + a (eⁱβ − eⁱα)]* = [b (eⁱφ − eⁱθ)] [b (eⁱφ − eⁱθ)]*
      [a + a (eⁱβ − eⁱα)] [a + a (e^-iβ − e^-iα)] = b ² (eⁱφ − eⁱθ) (e^-iφ − e^-iθ)
      a ² + a ² (eⁱβ − eⁱα) (e^-iβ − e^-iα)   +   a ² (eⁱβ − eⁱα + e^-iβ − e^-iα)   =   a c (eⁱφ − eⁱθ) (e^-iφ − e^-iθ)
      a + a (1 − eⁱ(α-β) − e^-i(α-β) + 1)   +   a (eⁱβ + e^-iβ − eⁱα − e^-iα)   =   c (1 − eⁱ(θ-φ) − e^-i(θ-φ) + 1)
      a + 2 a (1 − cos(α − β))   +   2 a (cos β − cos α)   =   2 c (1 − cos(θ − φ))

Subtract our previous equation for 0 from the left-hand side
      2 c (1 − cos(α − β))   =   2 c (1 − cos(θ − φ))
      α − β   =   θ − φ

Return to our two alternative expressions for P:
      P = ½ a + a eⁱβ + b eⁱφ
      P = ½ a + eⁱβ (a + b eⁱ(φ-β)) = ½ a + eⁱβ (a + b eⁱ(θ-α))
Also:
      P = − ½ a + a eⁱα + b eⁱθ = − ½ a + eⁱα (a + b eⁱ(θ-α))
We can eliminate the bracketed expression, and so eliminate b:
      P eⁱα = ½ a eⁱα + eⁱα eⁱβ (a + b eⁱ(θ-α))
      P eⁱβ = − ½ a eⁱβ + eⁱβ eⁱα (a + b eⁱ(θ-α))
      P (eⁱα − eⁱβ) = a (eⁱα + eⁱβ)
Multiply each side by (e^-iα − e^-iβ) in orer to make the LHS real:
      P (eⁱα − eⁱβ) (e^-iα − e^-iβ) = a (eⁱα + eⁱβ) (e^-iα − e^-iβ)
      P (1 − eⁱ(α-β) − e^-i(α-β) + 1) = a (1 − eⁱ(α-β) + e^-i(α-β) − 1)
      P (2 − 2 cos(α−β)) = − a (eⁱ(α-β) − e^-i(α-β)) = − 2 i a sin(α−β)
Thus the real part of P is zero — Q.E.D.