Let $\jmath$ be the isometry so that $\jmath(ABCD)=BADC$. Let $\jmath(M)=T$. Then we need to prove $TB \cdot AN + AT \cdot BN + DT \cdot CN \geq CT \cdot DN$. Then by Ptolemy $TB \cdot AN + AT \cdot BN \geq TN \cdot AB=TN \cdot CD$. Again, by Ptolemy $TN \cdot CD + DT \cdot CN \geq CT \cdot DN$. Finally $TB \cdot AN + AT \cdot BN + DT \cdot CN \geq TN \cdot CD +DT \cdot CN \geq CT \cdot DN$ $Q.E.D$

What is Isometry?

Beautiful solution,congratulations pajaraja

junior2001 wrote: What is Isometry? $j$ is isometry if for all $X$ and $Y$ we have $XY=X'Y'$, where $j(X)=X'$ and $j(Y)=Y'$.

Pajaraja wrote: Let $\jmath$ be the isometry so that $\jmath(ABCD)=BADC$. Let $\jmath(M)=T$. Then we need to prove $TB \cdot AN + AT \cdot BN + DT \cdot CN \geq CT \cdot DN$. Then by Ptolemy $TB \cdot AN + AT \cdot BN \geq TN \cdot AB=TN \cdot CD$. Again, by Ptolemy $TN \cdot CD + DT \cdot CN \geq CT \cdot DN$. Finally $TB \cdot AN + AT \cdot BN + DT \cdot CN \geq TN \cdot CD +DT \cdot CN \geq CT \cdot DN$ $Q.E.D$ Does there exists an isometry $\jmath$ such that $\jmath(ABCD)=(BADC)$?

Actually,for every two congurent figures A and B in Euclidean space there exists an isometry that sends A to B.In this case isometry is axial reflection over the line through midpoints of AB and CD.