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 $(\overrightarrow{\rm EC}, \overrightarrow{\rm ED})\equiv 2(SC, SB) \equiv 2(SA, SD)$

Application:

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

1. A Proof of the Napoleon theorem: Apply the theorem part one with $\beta=120^0$ and $\alpha=30^0$ (See Figure 1).

2. A Proof of the Bottema theorem: Apply the theorem part one with $\beta=45^0$ and $\alpha=90^0$ (See Figure 1).

3. A Proof of the Brahmagupta theorem: Apply the theorem part two with $\gamma=90^0$ (See Figure 1).

See also:

• Relative a generalization Bottema theorem