---
title: "Midpoints and Parallels"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 59
exercises: 9
source: https://one-course.com/books/math/1/en/chapter/59-midpoints-and-parallels
---

# Chapter 59 — Midpoints and Parallels

Join the midpoints of two sides of a triangle: the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) you get is always [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the third side, and exactly [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) as long. This “midpoint theorem” and its converse are the first taste of a great idea — [parallels](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) cut lengths proportionally — which blossoms into Thales’ theorem in [Chapter 68](https://one-course.com/books/math/1/en/chapter/68-thales-theorem#ch-g9-thales).

## 59.1 The midpoint theorem

**Theorem 59.1 (Midpoint theorem).**

In a triangle $ABC$, let $I$ be the midpoint of $[AB]$ and $J$ the midpoint of $[AC]$. Then the line $(IJ)$ is [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(BC)$, and

$$
IJ = \frac{BC}{2} .
$$

**Idea of proof.** Let $K$ be the [symmetric](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-def) of $I$ about $J$ (a half-turn, [Chapter 52](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#ch-g7-central)). The [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) $AICK$ has diagonals $[IK]$ and $[AC]$ crossing at their common midpoint $J$: it is a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram), so $KC$ is [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $AI$, i.e. to $(AB)$, and $KC = AI = IB$. Then $IBCK$ has two opposite sides ($[IB]$ and $[KC]$) [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) and equal: it is a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) too ([Theorem 52.9](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#thm-g7-central-recognize)), so $(IK) \parallel (BC)$ — that is $(IJ) \parallel (BC)$ — and $BC = IK = 2\,IJ$. ∎

![The midpoint segment (IJ) (red) is parallel to the third side (BC) and half as long.](https://one-course.com/images/onecourse/chapters/math-1/g8-midpoints/fig-c5593cfefe80.svg)

*The midpoint [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[IJ]$ (red) is [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the third side $(BC)$ and [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) as long.*

**Example 59.2.**

In a triangle $ABC$ with $BC = 9$ cm, the midpoints $I$ of $[AB]$ and $J$ of $[AC]$ are joined. Without any measurement: $(IJ) \parallel (BC)$ and $IJ = \frac92 = 4.5$ cm. The small triangle $AIJ$ is a half-size copy of $ABC$ — its [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) is [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) of $ABC$ as well.

**Theorem 59.3 (Converse).**

In a triangle $ABC$, the line through the midpoint $I$ of $[AB]$ and *[parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)* to $(BC)$ crosses the side $[AC]$ at its midpoint.

**Proof.** *Admitted at this level.* ∎

**Example 59.4.**

A path crosses a triangular field, starting at the middle of one edge and running [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the opposite edge. The converse guarantees that the path exits exactly at the middle of the other edge — no measuring needed on the far side.

**Method 59.5 (Choosing between theorem and converse).**

1. *Two midpoints known* $\Rightarrow$ use the direct theorem: conclude parallelism and half-length.
2. *One midpoint and a [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) known* $\Rightarrow$ use the converse: conclude that the crossing point is a midpoint.
3. Write which triangle, which midpoints and which theorem you use — one sentence each.

## 59.2 Half-size triangles

**Proposition 59.6 (The midpoint triangle).**

Joining the three midpoints of the sides of a triangle cuts it into four triangles of equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area), each a half-size copy of the original.

**Proof.** Each side of the midpoint triangle is [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to a side of $ABC$ and [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) as long ([Theorem 59.1](#thm-g8-midpoints-direct), applied three times). The four small triangles have sides of the same three lengths, so they are identical copies; together they tile $ABC$, so each has a [quarter](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of its [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area). ∎

![The midpoint triangle IJK cuts ABC into four equal triangles — a picture worth remembering.](https://one-course.com/images/onecourse/chapters/math-1/g8-midpoints/fig-b3c1cb0867de.svg)

*The midpoint triangle $IJK$ cuts $ABC$ into four equal triangles — a picture worth remembering.*

**Remark 59.7 (Towards Thales).**

The midpoint theorem is the case “one [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half)” of a more general fact: a line [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to one side of a triangle cuts the two other sides in *equal ratios* — one third and one third, two fifths and two fifths … That general statement is Thales’ theorem ([Chapter 68](https://one-course.com/books/math/1/en/chapter/68-thales-theorem#ch-g9-thales)); the midpoint case is the only one needed this year.

## 59.3 Exercises

**Exercise 59.1 ★.**

In a triangle $RST$, $M$ is the midpoint of $[RS]$ and $N$ the midpoint of $[RT]$, with $ST = 7$ cm. What can be said of the line $(MN)$ and the length $MN$? Cite the theorem used.

**Solution of Exercise 59.1.**

$M$ and $N$ are the midpoints of two sides of the triangle $RST$: by the midpoint theorem ([Theorem 59.1](#thm-g8-midpoints-direct)), $(MN) \parallel (ST)$ and $MN = \frac{ST}{2} = 3.5$ cm.

**Exercise 59.2 ★.**

In a triangle $ABC$, $I$ is the midpoint of $[AB]$ and the line through $I$ [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(BC)$ cuts $[AC]$ at $J$, with $AC = 11$ cm. Compute $AJ$, citing the theorem used.

**Solution of Exercise 59.2.**

$I$ is a midpoint and $(IJ) \parallel (BC)$: by the *converse* ([Theorem 59.3](#thm-g8-midpoints-converse)), $J$ is the midpoint of $[AC]$, so $AJ = \frac{11}{2} = 5.5$ cm.

**Exercise 59.3 ★.**

Draw a triangle $ABC$ with $AB = 8$ cm, $AC = 6$ cm, $BC = 7$ cm, place the midpoints $I$ of $[AB]$ and $J$ of $[AC]$, and measure $IJ$ to check the theorem. Compute the [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) of $AIJ$ without measuring anything else.

**Solution of Exercise 59.3.**

The measurement gives $IJ = 3.5$ cm $= \frac{BC}{2}$. [Perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) of $AIJ$: $AI = 4$, $AJ = 3$, $IJ = 3.5$: total $10.5$ cm — [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) of $ABC$ ($8 + 6 + 7 = 21$ cm).

**Exercise 59.4 ★.**

$IJK$ is the midpoint triangle of $ABC$, whose sides measure $10$, $12$ and $16$ cm. Give the three sides of $IJK$ and its [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter). How do the two [perimeters](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) compare?

**Solution of Exercise 59.4.**

Each side of $IJK$ is [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) a side of $ABC$: $5$, $6$ and $8$ cm. [Perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter): $19$ cm, [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of $10 + 12 + 16 = 38$ cm.

**Exercise 59.5 ★.**

The [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of a triangle is $36$ cm$^2$. What is the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of its midpoint triangle? Of each of the four small triangles?

**Solution of Exercise 59.5.**

The midpoint triangle is one of the four equal triangles of [Proposition 59.6](#prop-g8-midpoints-medial): [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $\frac{36}{4} = 9$ cm$^2$ — and each of the four small triangles has [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $9$ cm$^2$.

**Exercise 59.6 ★★.**

In a triangle $ABC$, $I$ is the midpoint of $[AB]$, $J$ that of $[AC]$ and $K$ that of $[BC]$. Show that $IJKB$ is a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram). (Compare $[IJ]$ and $[BK]$: [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)? equal?)

**Solution of Exercise 59.6.**

By the midpoint theorem in $ABC$: $(IJ) \parallel (BC)$, i.e. $(IJ) \parallel (BK)$, and $IJ = \frac{BC}{2} = BK$ (as $K$ is the midpoint of $[BC]$). Two opposite sides of $IJKB$ are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) and equal: it is a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) ([Theorem 52.9](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#thm-g7-central-recognize), criterion 3).

**Exercise 59.7 ★★.**

A triangular sail has its lowest edge $6.4$ m long. A reinforcement seam joins the midpoints of the two other edges. What length of seam is needed? What if the seam instead joins the points located at one [quarter](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of each edge, starting from the top [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def)? (Conjecture with a drawing; the proof is Thales’, [Chapter 68](https://one-course.com/books/math/1/en/chapter/68-thales-theorem#ch-g9-thales).)

**Solution of Exercise 59.7.**

The seam joins two midpoints: $\frac{6.4}{2} = 3.2$ m. At one [quarter](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) from the top [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def), the drawing suggests a seam of $\frac{6.4}{4} = 1.6$ m, [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the base — confirmed by Thales in [Chapter 68](https://one-course.com/books/math/1/en/chapter/68-thales-theorem#ch-g9-thales).

**Exercise 59.8 ★★.**

In triangle $ABC$, [right-angled](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) at $A$, $M$ is the midpoint of the hypotenuse $[BC]$, and the [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(AB)$ through $M$ meets $[AC]$ at $N$.

1. Show that $N$ is the midpoint of $[AC]$ .
2. Explain why $(MN)$ is [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(AC)$ .

**Solution of Exercise 59.8.**

*1.* $M$ is the midpoint of $[BC]$ and $(MN) \parallel (AB)$: by the converse of the midpoint theorem (in the triangle $ABC$, side $[AC]$), $N$ is the midpoint of $[AC]$.

*2.* $(MN) \parallel (AB)$ and $(AB) \perp (AC)$ ([right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) at $A$): a line [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(AB)$ is also [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(AC)$ ([Proposition 40.3](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#prop-g6-lines-facts)). So $(MN) \perp (AC)$ — the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[MN]$ is the [perpendicular bisector](https://one-course.com/books/math/1/en/chapter/42-axial-symmetry#def-g6-symmetry-bisector) piece showing again that $MA = MC$ ([Theorem 60.1](https://one-course.com/books/math/1/en/chapter/60-the-right-triangle-and-the-cosine#thm-g8-cosine-circle)).

**Exercise 59.9 ★★★.**

Let $ABCD$ be *any* [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon), and $I$, $J$, $K$, $L$ the midpoints of $[AB]$, $[BC]$, $[CD]$, $[DA]$ (the conjecture of [Exercise 52.11](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#exo-g7-central-11)).

1. Apply the midpoint theorem in the triangle $ABC$ to the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[IJ]$ , and in the triangle $ACD$ to the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[LK]$ : compare each to the diagonal $[AC]$ .
2. Conclude that $IJKL$ is always a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) .

**Solution of Exercise 59.9.**

*1.* In triangle $ABC$, $I$ and $J$ are midpoints of $[AB]$ and $[BC]$: $(IJ) \parallel (AC)$ and $IJ = \frac{AC}{2}$. In triangle $ACD$, $L$ and $K$ are midpoints of $[DA]$ and $[CD]$: $(LK) \parallel (AC)$ and $LK = \frac{AC}{2}$.

*2.* So $[IJ]$ and $[LK]$ are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) (both [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the diagonal $(AC)$) and equal (both [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) of $AC$): $IJKL$ is a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) ([Theorem 52.9](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#thm-g7-central-recognize), criterion 3) — for *every* [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) $ABCD$, as conjectured in [Exercise 52.11](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#exo-g7-central-11).

## 59.4 Problem: The three medians and the center of gravity

**Problem 59.1.**

Weekend problem — the medians of a triangle meet at a single point, two thirds of the way down each of them

Cut a triangle out of stiff cardboard and it will balance, perfectly flat, on a pencil tip placed at one special point: the *center of gravity* of the triangle. This problem finds that point. A *median* of a triangle is the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) joining a [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) to the *midpoint* of the opposite side. You will prove — with the midpoint theorem used twice, in an unexpected way — that the three medians all pass through one point, and locate it precisely.

**Part I — Warm-up.**

1. Draw a large triangle $ABC$ (no special shape), construct the midpoints $I$ of $[AB]$ , $J$ of $[AC]$ and $K$ of $[BC]$ , and draw the three medians $[AK]$ , $[BJ]$ , $[CI]$ . What do you observe?
2. Prove that a median cuts the triangle into two triangles of equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) . (Compare bases and heights, and use the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) formula: base $\times$ height $\div\ 2$ .) For a triangle of [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $36$ cm $^2$ , what are the two pieces’ [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) ?
3. In the triangle $ABC$ , what does [Theorem 59.1](#thm-g8-midpoints-direct) say about the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[IJ]$ ?

**Part II — Two medians meet two thirds of the way.** The medians $[BJ]$ and $[CI]$ cross at a point; call it $G$. Let $M$ be the midpoint of $[GB]$ and $N$ the midpoint of $[GC]$.

4. Apply the midpoint theorem *in the triangle $GBC$* : what does it say about the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[MN]$ ?
5. Deduce that $(IJ) \parallel (MN)$ and $IJ = MN$ , and conclude with [Theorem 52.9](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#thm-g7-central-recognize) that $IJNM$ is a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) .
6. Explain why $I$ and $N$ both lie on the median $(CI)$ , and $J$ and $M$ both on the median $(BJ)$ — so the diagonals of the [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) $IJNM$ are $[IN]$ and $[JM]$ , and they cross exactly at $G$ . What does the *definition* of a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) then say about $G$ ?
7. Deduce that $BM = MG = GJ$, and conclude: $$BG = 2 \times GJ$$ — the point $G$ sits on the median $[BJ]$ at two thirds of the way from the [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) $B$. State and justify the analogous result on the median $[CI]$.
8. Now forget $[CI]$ and run the very same argument on the pair of medians $[BJ]$ and $[AK]$ : their crossing point also sits at two thirds of $[BJ]$ from $B$ . Explain why this forces it to be the *same* point $G$ — and conclude that all three medians pass through $G$ .
9. In a triangle where the median $[BJ]$ measures $9$ cm, how far is $G$ from $B$ , and from $J$ ?
10. Justify (one sentence) that $AG = 2 \times GK$ as well.

**Part III — Six equal slices, and coordinates.** The three medians cut the triangle into six small triangles around $G$.

11. In the triangle $GBC$ , the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[GK]$ is a median. Deduce that the two small triangles $GBK$ and $GKC$ have equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) .
12. Using question 2 in the triangles $ABK$ and $AKC$ (both cut by the median $[AK]$ of $ABC$ ), show that the triangles $ABG$ and $ACG$ have equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) . Repeating with another median, conclude that the three triangles $ABG$ , $BCG$ , $CAG$ all have [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) one third of $ABC$ .
13. Each of these three triangles is cut in [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) by a piece of median (for instance $[GI]$ is a median of the triangle $ABG$ — seen from which [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) ?). Conclude: the six small slices around $G$ all have the same [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) , one sixth of the triangle.
14. On a coordinate grid, plot $A(0, 0)$, $B(6, 0)$ and $C(0, 6)$, with $K(3, 3)$ the midpoint of $[BC]$. Use the two-thirds theorem of Part II to compute the coordinates of $G$ on the median $[AK]$; then check that the same point sits at two thirds of the median $[BJ]$, where $J(0, 3)$. Compare $G$’s coordinates with the averages $$\frac{0 + 6 + 0}{3},  \qquad  \frac{0 + 0 + 6}{3} .$$
15. The balance finale: using questions 2 and 8, explain why the cardboard triangle balances on a knife edge laid along any median — equal amounts of cardboard on each side — and why the pencil tip must therefore be placed at $G$ , the point the physicists call the [center of gravity](#pb-g8-midpoints-1) . (A fully rigorous balance proof needs the integral calculus of the university [volumes](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume) ; the equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) make it believable today.)

**Solution of Problem 59.1.**

**1.** Whatever triangle is drawn, the three medians appear to pass through a single point, situated inside the triangle, visibly closer to each side’s midpoint than to the opposite [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def). The rest of the problem proves it.

**2.** Take the median $[AK]$ of the triangle $ABC$: the triangles $ABK$ and $AKC$ have bases $BK$ and $KC$ of the same length ($K$ is the midpoint of $[BC]$) carried by the same line $(BC)$, and the same height: the distance from $A$ to $(BC)$. By the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) formula, each has [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) base $\times$ height $\div\ 2$, with equal bases and equal heights: the [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) are equal. For a triangle of [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $36$ cm$^2$: two pieces of $18$ cm$^2$ each.

**3.** $I$ and $J$ are the midpoints of the sides $[AB]$ and $[AC]$ of the triangle $ABC$, so by [Theorem 59.1](#thm-g8-midpoints-direct): $(IJ) \parallel (BC)$ and $IJ = \frac{BC}{2}$.

**4.** $M$ and $N$ are the midpoints of the sides $[GB]$ and $[GC]$ of the triangle $GBC$, so by the same theorem: $(MN) \parallel (BC)$ and $MN = \frac{BC}{2}$.

**5.** Both $(IJ)$ and $(MN)$ are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(BC)$, hence [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to each other; and $IJ = \frac{BC}{2} = MN$. The [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) $IJNM$ thus has two opposite sides, $[IJ]$ and $[NM]$, [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) and of the same length: by criterion 3 of [Theorem 52.9](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#thm-g7-central-recognize), $IJNM$ is a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram).

**6.** $G$ lies on the median $[CI]$, and $N$ is the midpoint of $[GC]$, a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) carried by that same median line: so $I$, $N$ (and $G$, $C$) all lie on $(CI)$. Likewise $J$, $M$ and $G$ lie on the median line $(BJ)$. In the [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) $IJNM$, the diagonals are $[IN]$ and $[JM]$; they are carried by the two median lines, which cross at $G$ — so the diagonals cross at $G$. By the *definition* of a [parallelogram](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#def-g7-central-parallelogram) (criterion 1 of [Theorem 52.9](https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms#thm-g7-central-recognize): the diagonals have the same midpoint), $G$ is the midpoint of $[IN]$ *and* of $[JM]$: $GI = GN$ and $GJ = GM$.

**7.** $M$ is the midpoint of $[GB]$, so $BM = MG$; and $MG = GJ$ by question 6. Hence $BM = MG = GJ$: the median piece $[BJ]$ is cut into three equal parts, of which $BG$ takes two:

$$
BG = 2 \times GJ .
$$

Symmetrically, $N$ is the midpoint of $[GC]$ and $GN = GI$, so $CN = NG = GI$ and $CG = 2 \times GI$: on the median $[CI]$ too, the crossing point sits two thirds of the way from the [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def).

**8.** The argument used nothing special about the pair $[BJ]$, $[CI]$: run on the pair $[BJ]$, $[AK]$, it shows their crossing point also lies on $[BJ]$ at two thirds of the way from $B$. But there is only *one* point of the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[BJ]$ at distance $\frac23 BJ$ from $B$ — so the crossing point of $[BJ]$ and $[AK]$ is the very same $G$. All three medians therefore pass through $G$: they are *concurrent*.

**9.** $BG = \frac23 \times 9 = 6$ cm and $GJ = \frac13 \times 9 = 3$ cm.

**10.** By question 8 the same two-thirds computation holds on the median $[AK]$ (run the argument of questions 4–7 with the pair $[AK]$, $[BJ]$): $AG = 2 \times GK$.

**11.** In the triangle $GBC$, the point $K$ is the midpoint of the side $[BC]$, so $[GK]$ is a median of $GBC$; by question 2, the triangles $GBK$ and $GKC$ have equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area).

**12.** The median $[AK]$ cuts $ABC$ into $ABK$ and $AKC$ of equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) (question 2). Removing from each its small triangle of question 11:

$$
\text{area}(ABG) = \text{area}(ABK) - \text{area}(GBK),
\qquad
\text{area}(ACG) = \text{area}(AKC) - \text{area}(GKC),
$$

and the removed [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) are equal, so $\text{area}(ABG) = \text{area}(ACG)$. The same computation along the median $[BJ]$ — which halves $ABC$ into $ABJ$ and $CBJ$, while $[GJ]$, a median of the triangle $GAC$, halves it into $GAJ$ and $GCJ$ — gives $\text{area}(ABG) = \text{area}(CBG)$. So the three triangles $ABG$, $BCG$, $CAG$ have equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area), and since together they tile $ABC$, each has [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) one third of the whole.

**13.** In the triangle $ABG$, the point $I$ is the midpoint of the side $[AB]$, so $[GI]$ is a median of $ABG$ *from the [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) $G$*: it cuts $ABG$ into two triangles of equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) (question 2), namely $AIG$ and $IBG$. The same happens in $BCG$ (median $[GK]$) and in $CAG$ (median $[GJ]$). Each third of $ABC$ is cut in [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half): the six slices around $G$ each have [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) $\frac13 \times \frac12 = \frac16$ of the triangle.

**14.** On the median $[AK]$, from $A(0,0)$ to $K(3,3)$: $G$ sits two thirds of the way, so its coordinates are $\left(\frac23 \times 3,\ \frac23 \times 3\right) = (2, 2)$. On the median $[BJ]$, from $B(6, 0)$ to $J(0, 3)$: two thirds of the way means adding two thirds of the displacement ($-6$ horizontally, $+3$ vertically):

$$
\left(6 + \tfrac23 \times (-6),\ 0 + \tfrac23 \times 3\right)
= (2, 2) .
$$

The same point — as question 8 promised. And $\frac{0 + 6 + 0}{3} = 2$, $\frac{0 + 0 + 6}{3} = 2$: the [center of gravity](#pb-g8-midpoints-1)’s coordinates are the *averages* of the three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def)’ coordinates.

**15.** By question 2, a knife edge laid along a median has the same amount of cardboard on each side (equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area)), so the triangle balances along each of the three median lines. A point of balance must lie on all three balance lines at once, and by question 8 the three medians share exactly one point: $G$. That is where the pencil tip goes. (Equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) on both sides is the believable version of the argument; the honest one, weighing *where* the cardboard sits and not just how much there is, needs the integral calculus of the university [volumes](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-volume) — which confirms $G$.)