Possible new theorem in plane geometry encompassing 5 famous geometry theorems

Question

I am looking for a proof of a generalization Napoleon theorem, Bottema theorem and Brahmagupta theorem and van Aubel theorem, and Finsler–Hadwiger theorem in one configuration, as follows:

Let four points $A, B, C, D$ in the plain, the perpendicular bisector of $AB$ meets the perpendicular bisector 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 $\beta=120^\circ$ and $\alpha=30^\circ$ (See Figure 1).

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

3. A proof of Van Aubel theorem. Apply the theorem part one, with $\alpha=\gamma=45^\circ$

4. A roof of Finsler–Hadwiger theorem. Apply the theorem part one, with $\alpha=\gamma=45^\circ$

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

See also:

• Relative a generalization Bottema theorem

Answer 1

Let $O$ be a circumcentre of $\triangle ADS$. Then $\triangle OAS \sim \triangle PAB$, thus $\triangle POA\sim \triangle BSA$ and $BS:PO=SA:OA$, $\angle (BS,PO)=\angle (SA,OA)$. Analogously $CS:PO=SD:OD$ and $\angle (CS,PO)=\angle (SD,OD)$. This gives that $BS:CS=SA:SD$ and by angle chasing $\angle BSC=\gamma$, so $\triangle BSC\sim \triangle ASD$ (with different orientation), and we get the first part of your question.

Now let $K$ by symmetric to $C$ with respect to line $BS$. Let $E$ be a point such that $\triangle CED\sim \triangle CSK$, so that $E$ lies on the perpendicular bisector to $CD$ and $\angle CED=2\gamma$. We need to prove that $SE\perp AB$. We get $\triangle CES\sim \triangle CDK$, thus $\angle (ES,DK)=\angle (CE,CD)=\pi/2-\gamma$. But $\triangle DSK\sim \triangle ASB$, thus $\angle (DK,AB)=\angle (DS,AS)=\gamma$, and $\angle (ES,AB)=\angle (ES,DK)+\angle (DK,AB)=\pi/2$.