$G$ is the centroid of $\triangle\mathit{ABC}$. $X$, $Y$, and $Z$ are points on the sides (or lines through the sides) $\overline{\mathit{BC}}$, $\overline{\mathit{AC}}$, and $\overline{\mathit{AB}}$, respectively, such that $G$, $X$, $Y$, and $Z$ are collinear. If $Y$ is between $G$ and $Z$, \begin{equation*} \frac{1}{\mathit{GX}} = \frac{1}{\mathit{GY}} + \frac{1}{\mathit{GZ}} . \end{equation*}

How can I show this without using vectors (directed line segments) and Menelaus' Theorem?

A solution that I have seen assumes that X is a point on $\overline{\mathit{BC}}$ and constructs points U and V that trisect the side; U is half as far from B as it is from C. (We assume that X is distinct from both U or V so that the line through G and X intersects the other two sides.) Since G is half as far from the endpoint of the median on $\overline{\mathit{BC}}$ as it is from A, and since U is half as far from B as is $\overline{\mathit{GU}}$ is parallel to side $\overline{\mathit{AB}}$. Likewise, $\overline{\mathit{GV}}$ is parallel to side $\overline{\mathit{AC}}$. \begin{equation*} \frac{GX}{GZ} = \frac{UX}{BU} \qquad \text{and} \qquad \frac{GX}{GY} = \frac{VX}{CV} . \end{equation*} That is fine.

Now the solution says that $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$ so that \begin{equation*} \frac{\mathit{GX}}{\mathit{GZ}} = \frac{\mathit{UX}}{\mathit{BU}} = \frac{\mathit{UZ} - \mathit{XZ}}{\mathit{BU}} \end{equation*} and \begin{equation*} \frac{\mathit{UZ}}{\mathit{BU}} = \frac{\mathit{GX}}{\mathit{GZ}} + \frac{\mathit{XZ}}{\mathit{BU}} . \end{equation*} Likewise, since $\mathit{VX} = \mathit{VZ} - \mathit{XZ}$, \begin{equation*} \frac{\mathit{VZ}}{\mathit{CV}} = \frac{\mathit{GX}}{\mathit{GY}} + \frac{\mathit{XZ}}{\mathit{CV}} . \end{equation*}

I would appreciate an elementary explanation of these last two equalities - involving only similar triangles and proportions - without mentioning $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$ and $\mathit{VX} = \mathit{VZ} - \mathit{XZ}$.

  • $\begingroup$ The first of the last two equalities is simply a rearrangement of the terms from the equation preceding (after separating the last fraction into $UZ/BU-XZ/BU$), and that follows directly from the first proportion you say is "fine". So ... What more explanation do you need? $\endgroup$ – Blue Jul 17 at 20:16
  • $\begingroup$ I should edit my post. These equalities are based on the equality $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$. $\endgroup$ – A gal named Desire Jul 17 at 21:20
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    $\begingroup$ Can I get these last two equalities while avoiding mentioning the equality $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$? $\endgroup$ – A gal named Desire Jul 17 at 21:35
  • $\begingroup$ Post a picture also $\endgroup$ – Aqua Jul 19 at 10:44
  • $\begingroup$ @Aqua I have a TikZ diagram for it. I do not know how to upload it. $\endgroup$ – A gal named Desire Jul 19 at 13:33

Let $Z$ be placed on the line $AB$ such that $A$ is placed between $Z$ and $B$, $X$ be placed on the side $BC$,

$Y$ be placed on the side $AC$, $BD$ be a median of $\Delta ABC$, $GX=x$, $GY=y$ and $GZ=z$.

Thus, $$[Z,Y,G,X]=[A,Y,D,C],$$ which says $$\frac{ZG}{ZX}:\frac{YG}{YX}=\frac{AD}{AC}:\frac{YD}{YC}$$ or $$\frac{z}{x+z}\cdot\frac{x+y}{y}=\frac{\frac{AC}{2}}{AC}\cdot\frac{YC}{YC-\frac{1}{2}AC}$$ or $$\frac{(x+y)z}{(x+z)y}=\frac{1}{2-\frac{AC}{YC}}.$$ Now, let $GE||BC$ and $E\in DC$.

Thus, $$\frac{EC}{CY}=\frac{GX}{XY}$$ or $$\frac{\frac{1}{3}AC}{CY}=\frac{x}{x+y},$$ which says $$\frac{AC}{CY}=\frac{3x}{x+y}.$$ Id est, $$\frac{(x+y)z}{(x+z)y}=\frac{1}{2-\frac{3x}{x+y}}$$ or $$\frac{z}{(x+z)y}=\frac{1}{2y-x}$$ or $$2yz-xz=xy+yz$$ or $$\frac{1}{x}=\frac{1}{y}+\frac{1}{z}$$ and we are done!

Actually. By definition, for four points $A$, $B$, $C$ and $D$ which they are placed on the same line we have: $$[A,B,C,D]=\frac{AC}{AD}:\frac{BC}{BD}.$$ You can read about this here: https://en.wikipedia.org/wiki/Cross-ratio

  • $\begingroup$ $Y$ is a point on side $\overline{\mathit{AC}}$. So, $Y$ is between $A$ and $C$. $\endgroup$ – A gal named Desire Jul 22 at 16:14
  • $\begingroup$ You said "$C$ (is) placed between $A$ and $Y$." $\endgroup$ – A gal named Desire Jul 22 at 16:17
  • $\begingroup$ Also, you said "$[Z,G,Z,Y]=[A,D,C,Y]$." You must have a typographical error here. First, you have the point $Z$ twice in the expression on the left side of the equals sign. Second, you said $\overline{\mathit{BD}}$ is a median of the given triangle. That means $D$ is a point on side $\overline{\mathit{AC}}$. $Y$ is also a point on side $\overline{\mathit{AC}}$. So, all four points on the right side of the equals sign are collinear. $\endgroup$ – A gal named Desire Jul 22 at 16:23
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    $\begingroup$ I am driving now. Wait please, I'll fix and explain all $\endgroup$ – Michael Rozenberg Jul 22 at 16:32
  • $\begingroup$ @A gal named Desire I added something. See now. $\endgroup$ – Michael Rozenberg Jul 22 at 17:33

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