I am looking for a proof of Generalization Napoleon theorem, Bottema theorem and Brahmagupta theorem on one configuration as follows:

Let four points $A, B, C, D$ in the plain, the perpendicular of $AB$ meets the perpendicular of $CD$ at $P$. then always exist only one point $S$ such that:

1) $(\overrightarrow{\rm PD}, \overrightarrow{\rm PC})\equiv 2(AD, AS) \equiv 2(BS, BC)$ and $(\overrightarrow{\rm PB}, \overrightarrow{\rm PA})\equiv 2(CB, CS) \equiv 2(DS, DA)$ (in the figure). 

2) The line through $S$ meets the perpendicular bisector of $CD$ at $E$ then $ES \perp AB$ if only if $(\overrightarrow{\rm EC}, \overrightarrow{\rm ED})\equiv 2(SC, SB) \equiv 2(SA, SD)$ 

[![enter image description here][1]][1]

**Application:**

If the problem was proved. Then we can apply the theorem to proof two famous theorem:

1. A Proof of the [Napoleon theorem](https://en.wikipedia.org/wiki/Napoleon%27s_theorem): Apply the theorem part one with $\beta=120^0$ and $\alpha=30^0$ (See Figure 1).

2. A Proof of the [Bottema theorem](https://en.wikipedia.org/wiki/Bottema%27s_theorem): Apply the theorem part one with $\beta=45^0$ and $\alpha=90^0$ (See Figure 1).

3. A Proof of the [Brahmagupta theorem](https://en.wikipedia.org/wiki/Brahmagupta%27s_formula): Apply the theorem part two with $\gamma=90^0$ (See Figure 1).

**See also:**

* [Relative a generalization Bottema theorem](https://artofproblemsolving.com/community/q3h557966p3244119)


  [1]: https://i.sstatic.net/n93J2.png