# An equality from the line through the centroid of a triangle

$$G$$ 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 $$\mathit{VX} = \mathit{VZ} - \mathit{XZ}$$.

• 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? – Blue Jul 17 at 20:16
• I should edit my post. These equalities are based on the equality $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$. – A gal named Desire Jul 17 at 21:20
• Can I get these last two equalities while avoiding mentioning the equality $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$? – A gal named Desire Jul 17 at 21:35
• Post a picture also – Aqua Jul 19 at 10:44
• @Aqua I have a TikZ diagram for it. I do not know how to upload it. – 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

• $Y$ is a point on side $\overline{\mathit{AC}}$. So, $Y$ is between $A$ and $C$. – A gal named Desire Jul 22 at 16:14
• You said "$C$ (is) placed between $A$ and $Y$." – A gal named Desire Jul 22 at 16:17
• 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. – A gal named Desire Jul 22 at 16:23
• I am driving now. Wait please, I'll fix and explain all – Michael Rozenberg Jul 22 at 16:32
• @A gal named Desire I added something. See now. – Michael Rozenberg Jul 22 at 17:33