Prove that if the opposite sides of a skew (non-planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.
Problem
Source: 1977 USAMO Problem 4
Tags:
Luis González
04.04.2010 08:08
Label $ABCD$ the given quadrilateral. $M,N$ denote the midpoints of the diagonals $AC,BD,$ respectively. $\bullet$ Assume that $AD = CB$ and $AB = CD.$ Then $\triangle ADC \cong \triangle CBA$ by SSS criterion $\Longrightarrow$ their medians $DM$ and $BM$ are congruent. Hence $\triangle DMB$ is isosceles with apex $M.$ The median $MN$ is identical to the altitude on $DB$ $\Longrightarrow$ $MN \perp DB.$ Likewise, $\triangle ADB \cong \triangle CBD,$ then $\triangle ANC$ is isosceles with apex $N$ $\Longrightarrow$ $NM \perp AC.$ $\bullet$ Conversely, if $MN \perp DB$ and $NM \perp AC,$ the triangles $\triangle MDB$ and $\triangle NAC$ are isosceles with legs $MB = MD$ and $NA = NC,$ respectively, which implies that $CD^2 + CB^2 = AB^2 + AD^2 \ , \ AD^2 + CD^2 = AB^2 + CB^2.$ Substracting and adding both expressions yields $AD = CB$ and $AB = CD.$
huashiliao2020
15.04.2023 03:38
I finished the first part of the problem, by using SSS to show that triangles ADC and CBA are congruent, so if H and I are the midpoints of AC and BD, then DH=BH (CPCTC), with midpoint of BD as I, but then HI is perp. bisector of BD (even stronger), similarly with HI perp. bisector of AC.
However, I wasn't able to figure out the converse, could anyone give me some hints? EDit: @above above wrote: ... which implies that $CD^2 + CB^2 = AB^2 + AD^2 \ , \ AD^2 + CD^2 = AB^2 + CB^2.$ I didn't understand how you got there, is that some property of nonplanar quads? Maybe someone could clarify for me, since I understand I am bumping a thread where the post last sent was in 2010.