Primary & Middle School Mathematics · Grades 1–9

59Midpoints and Parallels

Join the midpoints of two sides of a triangle: the segment you get is always parallel to the third side, and exactly half as long. This “midpoint theorem” and its converse are the first taste of a great idea — parallels cut lengths proportionally — which blossoms into Thales’ theorem in Chapter 68.

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 to $(BC)$ , and

IJ = \frac{BC}{2} .

Idea of proof. Let $K$ be the symmetric of $I$ about $J$ (a half-turn, Chapter 52). The quadrilateral $AICK$ has diagonals $[IK]$ and $[AC]$ crossing at their common midpoint $J$ : it is a parallelogram, so $KC$ is parallel to $AI$ , i.e. to $(AB)$ , and $KC = AI = IB$ . Then $IBCK$ has two opposite sides ( $[IB]$ and $[KC]$ ) parallel and equal: it is a parallelogram too (Theorem 52.9), 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. — The midpoint segment $[IJ]$ (red) is parallel to the third side $(BC)$ and 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 is half the perimeter of $ABC$ as well.

Theorem 59.3 (Converse)

In a triangle $ABC$ , the line through the midpoint $I$ of $[AB]$ and parallel 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 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)

Two midpoints known $\Rightarrow$ use the direct theorem: conclude parallelism and half-length.
One midpoint and a parallel known $\Rightarrow$ use the converse: conclude that the crossing point is a midpoint.
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, each a half-size copy of the original.

Proof. Each side of the midpoint triangle is parallel to a side of $ABC$ and half as long (Theorem 59.1, 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 of its area. ∎

The midpoint triangle IJK cuts ABC into four equal triangles — a picture worth remembering. — 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” of a more general fact: a line parallel 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); 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

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), $(MN) \parallel (ST)$ 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 to $(BC)$ cuts $[AC]$ at $J$ , with $AC = 11$ cm. Compute $AJ$ , citing the theorem used.

Solution

Solution of Exercise 59.2.

$I$ is a midpoint and $(IJ) \parallel (BC)$ : by the converse (Theorem 59.3), $J$ 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 of $AIJ$ without measuring anything else.

Solution

Solution of Exercise 59.3.

The measurement gives $IJ = 3.5$ cm $= \frac{BC}{2}$ . Perimeter of $AIJ$ : $AI = 4$ , $AJ = 3$ , $IJ = 3.5$ : total $10.5$ cm — half the 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. How do the two perimeters compare?

Solution

Solution of Exercise 59.4.

Each side of $IJK$ is half a side of $ABC$ : $5$ , $6$ and $8$ cm. Perimeter: $19$ cm, half of $10 + 12 + 16 = 38$ cm.

Exercise 59.5 ★

The area of a triangle is $36$ cm $^2$ . What is the area of its midpoint triangle? Of each of the four small triangles?

Solution

Solution of Exercise 59.5.

The midpoint triangle is one of the four equal triangles of Proposition 59.6: area $\frac{36}{4} = 9$ cm $^2$ — and each of the four small triangles has 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. (Compare $[IJ]$ and $[BK]$ : parallel? equal?)

Solution

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 and equal: it is a parallelogram (Theorem 52.9, 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 of each edge, starting from the top vertex? (Conjecture with a drawing; the proof is Thales’, Chapter 68.)

Solution

Solution of Exercise 59.7.

The seam joins two midpoints: $\frac{6.4}{2} = 3.2$ m. At one quarter from the top vertex, the drawing suggests a seam of $\frac{6.4}{4} = 1.6$ m, parallel to the base — confirmed by Thales in Chapter 68.

Exercise 59.8 ★★

In triangle $ABC$ , right-angled at $A$ , $M$ is the midpoint of the hypotenuse $[BC]$ , and the parallel to $(AB)$ through $M$ meets $[AC]$ at $N$ .

Show that $N$ is the midpoint of $[AC]$ .
Explain why $(MN)$ is perpendicular to $(AC)$ .

Solution

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 at $A$ ): a line parallel to $(AB)$ is also perpendicular to $(AC)$ (Proposition 40.3). So $(MN) \perp (AC)$ — the segment $[MN]$ is the perpendicular bisector piece showing again that $MA = MC$ (Theorem 60.1).

Exercise 59.9 ★★★

Let $ABCD$ be any quadrilateral, and $I$ , $J$ , $K$ , $L$ the midpoints of $[AB]$ , $[BC]$ , $[CD]$ , $[DA]$ (the conjecture of Exercise 52.11).

Apply the midpoint theorem in the triangle $ABC$ to the segment $[IJ]$ , and in the triangle $ACD$ to the segment $[LK]$ : compare each to the diagonal $[AC]$ .
Conclude that $IJKL$ is always a parallelogram.

Solution

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 (both parallel to the diagonal $(AC)$ ) and equal (both half of $AC$ ): $IJKL$ is a parallelogram (Theorem 52.9, criterion 3) — for every quadrilateral $ABCD$ , as conjectured in Exercise 52.11.

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 joining a vertex 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.

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?
Prove that a median cuts the triangle into two triangles of equal areas. (Compare bases and heights, and use the area formula: base $\times$ height $\div\ 2$ .) For a triangle of area $36$ cm $^2$ , what are the two pieces’ areas?
In the triangle $ABC$ , what does Theorem 59.1 say about the segment $[IJ]$ ?

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 $[GC]$ .

Apply the midpoint theorem in the triangle $GBC$ : what does it say about the segment $[MN]$ ?
Deduce that $(IJ) \parallel (MN)$ and $IJ = MN$ , and conclude with Theorem 52.9 that $IJNM$ is a parallelogram.
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 $IJNM$ are $[IN]$ and $[JM]$ , and they cross exactly at $G$ . What does the definition of a parallelogram then say about $G$ ?
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 $B$ . State and justify the analogous result on the median $[CI]$ .
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 $G$ .
In a triangle where the median $[BJ]$ measures $9$ cm, how far is $G$ from $B$ , and from $J$ ?
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 $G$ .

In the triangle $GBC$ , the segment $[GK]$ is a median. Deduce that the two small triangles $GBK$ and $GKC$ have equal areas.
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. Repeating with another median, conclude that the three triangles $ABG$ , $BCG$ , $CAG$ all have area one third of $ABC$ .
Each of these three triangles is cut in half by a piece of median (for instance $[GI]$ is a median of the triangle $ABG$ — seen from which vertex?). Conclude: the six small slices around $G$ all have the same area, one sixth of the triangle.
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

$\frac{0 + 6 + 0}{3}, \qquad \frac{0 + 0 + 6}{3} .$
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. (A fully rigorous balance proof needs the integral calculus of the university volumes; the equal areas make it believable today.)

Solution

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. 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 formula, each has area base $\times$ height $\div\ 2$ , with equal bases and equal heights: the areas are equal. For a triangle of 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: $(IJ) \parallel (BC)$ 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 to $(BC)$ , hence parallel to each other; and $IJ = \frac{BC}{2} = MN$ . The quadrilateral $IJNM$ thus has two opposite sides, $[IJ]$ and $[NM]$ , parallel and of the same length: by criterion 3 of Theorem 52.9, $IJNM$ is a parallelogram.

6. $G$ lies on the median $[CI]$ , and $N$ is the midpoint of $[GC]$ , a segment 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 $IJNM$ , 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 (criterion 1 of Theorem 52.9: 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.

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 $[BJ]$ 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.

12. The median $[AK]$ cuts $ABC$ into $ABK$ and $AKC$ of equal areas (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 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, and since together they tile $ABC$ , each has 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 $G$ : it cuts $ABG$ into two triangles of equal areas (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: the six slices around $G$ each have 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’s coordinates are the averages of the three vertices’ coordinates.

15. By question 2, a knife edge laid along a median has the same amount of cardboard on each side (equal areas), 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 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 — which confirms $G$ .)