In this topic I want to share relation of the [Pythagorean theorem](http://mathworld.wolfram.com/PythagoreanTheorem.html), the [Stewart theorem](https://en.wikipedia.org/wiki/Stewart%27s_theorem) and the [British Flag theorem](https://en.wikipedia.org/wiki/British_flag_theorem), the [Apollonius' theorem](https://en.wikipedia.org/wiki/Apollonius%27_theorem) and the **Feuerbach-Luchterhand**. Since that I posed two generalizations of these theorems. **I. Relation of some Euclidean geometry theorems** 1. [Pythagorean theorem](http://mathworld.wolfram.com/PythagoreanTheorem.html): Let $ABC$ For a right triangle with legs $AB$ and $AC$ and hypotenuse $BC$ then: \begin{equation}AB^2+AC^2=BC^2 \end{equation} [![enter image description here][1]][1] 2. [Apollonius' theorem](https://en.wikipedia.org/wiki/Apollonius%27_theorem) in any triangle $ABC$, if $AD$ is a median, then \begin{equation}AB^2 + AC^2 = 2(AD^2+BD^2)\end{equation} [![enter image description here][3]][3] * Let $ABC$ be [Isosceles triangle](https://en.wikipedia.org/wiki/Isosceles_triangle) with $AB=AC$. Apply Apollonius' theorem we have $AB^2=AD^2+BD^2$, in this case $AD \perp BC$. This is the Pythagorean theorem with the right triangle $ABD$. So the Apollonius' theorem is a generalization of the Pythagorean theorem. 3. [Stewart's theorem](https://en.wikipedia.org/wiki/Stewart%27s_theorem) Let $A$, $B$, $C$ be points on a directed line $l$ in the Euclidean plane, and $P$ be a point anywhere in the plane. Then \begin{equation}PA^2.\overline{BC} + PB^2.\overline{CA} + PC^2.\overline{AB} + \overline{BC}.\overline{CA}.\overline{AB} = 0\end{equation} [![enter image description here][2]][2] * We let $B$ is midpoint of $AC$ we have: $\overline{BC}=\overline{AB}$, $\overline{CA}=-2\overline{AB}$. Since Stewart's theorem we have: \begin{equation}PA^2.\overline{AB} - 2PB^2.\overline{AB} + PC^2.\overline{AB} - 2.\overline{AB}.\overline{AB}.\overline{AB} = 0\end{equation} $\Leftrightarrow$ \begin{equation}PA^2 + PC^2 = 2(2PB^2+AB^2) = 0\end{equation} This is the Apollonius' theorem with the right triangle $PAC$ with median $PB$. So the Stewart's theorem is a generalization of the Apollonius' theorem. 4. [British flag theorem](https://en.wikipedia.org/wiki/British_flag_theorem) if a point ''P'' a plane of rectangle $ABCD$ then: \begin{equation}PA^2+PC^2=PD^2+PB^2\end{equation} [![enter image description here][4]][4] * Let $P \equiv D$ we have \begin{equation}DA^2+DC^2=DB^2\end{equation} $\Leftrightarrow$ \begin{equation}DA^2+DC^2=AC^2\end{equation} This is the Pythagorean theorem with the right triangle $DAC$. So the British flag theorem is a generalization of the Pythagorean theorem. **5. Feuerbach-Luchterhand** Let $ABCD$ be a cyclic quadrilateral, $P$ be a point on the plane then: \begin{equation}PA^2.DB.BC.CD-PB^2.AC.CD.DA+PC^2.BD.DA.AB-PD^2.CA.AB.BC = 0\end{equation} [![enter image description here][5]][5] * Let circles through $A, B, C, D$ is a line, and $D$ at infinity. Then $DB.CD=CD.DA=BD.DA=PD^2$. From the Feuerbach-Luchterhand we have: \begin{equation}PA^2.BC-PB^2.AC+PC^2.AB-CA.AB.BC = 0\end{equation} This is the Stewart theorem with three collinear points $A, B,C$ and $P$ on the plane. So the Feuerbach-Luchterhand is a generalization of the Stewart theorem. * Let cyclic quarilateral $ABCD$ is a rectangle. We have $AB=CD$ and $AD=BC$ and $AC=BD$. From the Feuerbach-Luchterhand we have: \begin{equation}PA^2.AC.AD.AB-PB^2.AC.AB.DA+PC^2.AC.DA.AB-PD^2.CA.AB.AD = 0\end{equation} $\Leftrightarrow$ \begin{equation}PA^2-PB^2+PC^2-PD^2 = 0\end{equation} This is the British flag theorem with rectangle $ABCD$ and $P$ on the plane. So the Feuerbach-Luchterhand is a generalization of the British flag theorem. **II. More generalization and I am looking for get a solution** We write Feuerbach-Luchterhand with a form following: Let $A_1A_2A_3A_4$ be a cyclic quadrilateral, $P$ be a point on the plane then: \begin{equation}PA_1^2.\frac{A_{4}A_{2}}{A_{1}A_{4}.A_{1}A_{2}}-PA_2^2.\frac{A_{1}A_{3}}{A_{2}A_{1}.A_{2}A_{3}}+PA_3^2.\frac{A_{2}A_{4}}{A_{3}A_{2}.A_{3}A_{4}}-PA_4^2.\frac{A_{3}A_{1}}{A_{4}A_{3}.A_{4}A_{1}}=0 \end{equation} I posed two generalization of Feuerbach-Luchterhand as follows: >* **Problem 1:**(First generalization) Let 2n-convex cyclic polygon $A_1A_2A_3...A_{2n}$, let $P$ be a point on the plane, then: \begin{equation} \sum_{i=1}^{2n} (-1)^{i+1}.PA_i^2.\frac{A_{i-1}A_{i+1}}{A_{i}A_{i-1}.A_{i}A_{i+1}}=0 \end{equation} > * **Problem 2:**(Second generalization) Let two direct similar 2n-convex cyclic polygon $A_1A_2A_3...A_{2n}$ and $B_1B_2B_3...B_{2n}$, then: \begin{equation} \sum_{i=1}^{2n} (-1)^{i+1}.B_iA_i^2.\frac{A_{i-1}A_{i+1}}{A_{i}A_{i-1}.A_{i}A_{i+1}}=0 \end{equation} Where $A_0=A_{2n}$ and $A_{2n+1}=A_1$ [1]: https://i.sstatic.net/dIugV.png [2]: https://i.sstatic.net/b0fny.png [3]: https://i.sstatic.net/Nt3xe.png [4]: https://i.sstatic.net/0HrKA.png [5]: https://i.sstatic.net/5ZvHr.png