Let $ABCD$ be a convex quadrilateral such that no two opposite sides are parellel to each other. Denote by $Q$ the intersection of lines $AD$ and $BC$ and by $R$ the intersection of lines $AB$ and $CD$. Let $X,Y,Z$ be midpoints of $AC, BD$ and $QR$ respectively. Prove that $X,Y,Z$ lie on the same line.

I am not getting any approach to solve this question. Please help.

Cycl Quad XYZ Concur

  • $\begingroup$ Can you provide a pic? It'll be better to visualise? $\endgroup$ – SchrodingersCat Jan 11 '16 at 17:29
  • $\begingroup$ The edges $AD$ and $BC$ shouldn't intersect, nor should $AB$ and $CD.$ Perhaps you should clarify what you mean by "convex" in this context. $\endgroup$ – Cameron Buie Jan 11 '16 at 17:30
  • $\begingroup$ @Cameron Buie not edges but lines $AD$ and $BC$ can intersect. I have corrected it. $\endgroup$ – Satvik Mashkaria Jan 11 '16 at 17:34
  • $\begingroup$ @ SchrodingersCat I have created a geogebra figure but I don't know how to put it here. $\endgroup$ – Satvik Mashkaria Jan 11 '16 at 17:38
  • $\begingroup$ @Cameron Buie convex quadrilateral is a quadrilateral such that line SEGMENT $AC$ and $BD$ intersect at one point. $\endgroup$ – Satvik Mashkaria Jan 11 '16 at 17:41

This is also known the Newton-Gauss line of $ABCD$. The usual proof is either by area considerations or by Menelaus' theorem. However, I present a purely synthetic approach.

Let $L,M,N$ be the midpoints of $\overline{AB},\overline{AR},\overline{BR}$. By considering midpoints in $\triangle ABC$, we have $XL\parallel BC$, and with $\triangle BQR$, we have $BQ\parallel NZ\implies XL\parallel NZ$. In a similar way, we can obtain $LY\parallel MZ$ and $MX\parallel NY$. But now Pappus' theorem on the hexagon $XMZNYL$ implies that $X,Y,Z$ are collinear.


It is well-known that in every complete quadrilateral, the midpoints of the three diagonals are collinear. See http://mathworld.wolfram.com/CompleteQuadrilateral.html


Here is an analytical solution. There are two constants $g_1,g_2$ such that $(1) \overrightarrow{QC}=g_1\overrightarrow{QB}$ and $(2) \overrightarrow{RA}=g_2\overrightarrow{RB}$. Denote by $L$ the intersection point of $(AD)$ with the parallel to $(DC)$ through $B$. Then $(3) \overrightarrow{QD}=g_1\overrightarrow{QL}$ (using Thales' theorem in triangle $QDC$) and $(4) \overrightarrow{DA}=g_2\overrightarrow{DL}$ (using Thales' theorem in triangle $ARD$).

Let us try to show that vectors $\overrightarrow{XZ}$ and $\overrightarrow{YZ}$ are collinear. Since $\overrightarrow{XZ}=\frac{1}{2}(\overrightarrow{CQ}+\overrightarrow{AR})= \frac{1}{2}(g_1\overrightarrow{BQ}+g_2\overrightarrow{BR})$ and $\overrightarrow{YZ}=\frac{1}{4}(\overrightarrow{BQ}+\overrightarrow{BR}+\overrightarrow{DQ}+\overrightarrow{DR})$, the idea is to express $\overrightarrow{DQ}+\overrightarrow{DR}$ in terms of $\overrightarrow{BQ}$ and $\overrightarrow{BR}$.

We will achieve this by finding two different relations satisfied by $\overrightarrow{DQ}$ and $\overrightarrow{DR}$, and then solving the system. First, note that $(5)\overrightarrow{DR}-\overrightarrow{DQ}=\overrightarrow{BR}-\overrightarrow{BQ}$. Next, using (3) in $g_1\overrightarrow{DL}=g_1\overrightarrow{DQ}+g_1\overrightarrow{QL}$ we deduce $(6)g_1\overrightarrow{DL}=(g_1-1)\overrightarrow{DQ}$. Also, using both (2) and (4) in $\overrightarrow{DR}=\overrightarrow{DA}+\overrightarrow{AR}$ we find that $\overrightarrow{DR}=g_2\big(\overrightarrow{DL}+\overrightarrow{BR}\big)$. Combining this last formula with (6), we obtain $(7) g_1\overrightarrow{DR}=g_2\big((g_1-1)\overrightarrow{DQ}+g_1\overrightarrow{BR}\big)$.

Now (5) and (7) provide us our $2\times 2$ linear system in $\overrightarrow{DQ}$ and $\overrightarrow{DR}$. Its unique solution is

$$ \overrightarrow{DQ}=\frac{g_1\overrightarrow{BQ}+(g_1g_2-g_1)\overrightarrow{BR}}{g_1+g_2-g_1g_2}, \ \overrightarrow{DR}=\frac{(g_1g_2-g_2)\overrightarrow{BQ}+g_2\overrightarrow{BR}}{g_1+g_2-g_1g_2}\tag{8} $$

This yields $\overrightarrow{YZ}=\frac{1}{2(g_1+g_2-g_1g_2)}\bigg(g_1\overrightarrow{BQ}+g_2\overrightarrow{BR}\bigg)$, so that $\overrightarrow{YZ}=\frac{1}{g_1+g_2-g_1g_2}\overrightarrow{XZ}$. This concludes the proof.


Here is a simple solution using coordinates:

Consider $A = (0, 0)$, $D = (D_x, 0)$ , $B = (B_x, B_y)$ and $C = (C_x, C_y)$.

We will get $X = \left( \dfrac {C_x} 2, \dfrac {C_y} 2 \right)$ and $Y = \left( \dfrac {B_x + D_x} 2, \dfrac {B_y} 2 \right)$. After some computation we will get the coordinates $Z$ in terms of $B_x, C_x, D_x$ and $B_y, C_y, D_y$. After some simplification, the solution becomes clear (I will not provide details unless OP requests so). Then, we can just check if they are on the same line and we are done!

  • $\begingroup$ This site uses Mathjax for typing . $\endgroup$ – Nizar Jan 15 '16 at 10:32
  • $\begingroup$ The reason for offering the bounty was that "The current answers do not contain enough detail". Do you think that your answer does? $\endgroup$ – Alex M. Jan 15 '16 at 11:13

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