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In this topic I want to share relation of the Pythagorean theorem, the Stewart theorem and the British Flag theorem, the Apollonius' theorem and the Feuerbach-Luchterhand. Since that I posed two conjectures of generalizations of these theorems. My question: I am looking for get a solution of my conjecture.

\begin{equation}PA^2 + PC^2 = 2(PB^2+AB^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.

In this topic I want to share relation of the Pythagorean theorem, the Stewart theorem and the British Flag theorem, the Apollonius' theorem, the Ptolemy's theorem and the Feuerbach-Luchterhand. Since that I posed two conjectures of generalizations of these theorems. My question: I am looking for get a solution of my conjecture.

\begin{equation}PA^2 + PC^2 = 2(PB^2+AB^2)\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.

• Let $P \equiv D$ then

\begin{equation}DA^2.DB.BC.CD-DB^2.AC.CD.DA+DC^2.BD.DA.AB-DD^2.CA.AB.BC = 0\end{equation}

$\Leftrightarrow$

\begin{equation}DA^2.DB.BC.CD-DB^2.AC.CD.DA+DC^2.BD.DA.AB= 0\end{equation}

$\Leftrightarrow$

\begin{equation}DA.BC-DB.AC+DC.AB= 0\end{equation}

$\Leftrightarrow$

\begin{equation}DB.AC = DA.BC+DC.AB\end{equation}

This is Ptolemy's theorem so Feuerbach-Luchterhand is a generalization of Ptolemy's theorem.

\begin{equation}PA^2 + PC^2 = 2(2PB^2+AB^2) = 0\end{equation}

\begin{equation}PA^2 + PC^2 = 2(PB^2+AB^2) = 0\end{equation}

• We let $B$ is midpoint of $AC$ we have: $\overline{BC}=\overline{AB}$, $\overline{CA}=-2\overline{AB}$. Since Stewart's theorem we have:

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 if a point ''P'' a plane of rectangle $ABCD$ then:

• Let $B$ is midpoint of $AC$ we have: $\overline{BC}=\overline{AB}$, $\overline{CA}=-2\overline{AB}$. Since Stewart's theorem we have:

This is the Apollonius' theorem with the triangle $PAC$ with median $PB$. So the Stewart's theorem is a generalization of the Apollonius' theorem.

4. British flag theorem if a point $P$ on the plane of rectangle $ABCD$ then: