Midsegments of triangles and trapezoids
Join the midpoints of two sides of a triangle. What does that new segment know about the third side?
Where you are 4 in Special segments in a triangle
Why this exists
Every side of a triangle has a midpoint, and finding it is easy: measure and halve, fold the paper, or average the coordinates. A median joins one of those midpoints to the opposite vertex. Join two of the midpoints to each other instead and something unexpected happens. The new segment knows the direction and the length of the third side, a side it never touches: it runs exactly parallel to it and is exactly half as long, whatever the shape of the triangle.
That makes two midpoints a cheap source of facts. A parallel line hands you equal angles, and a length you can measure hands you one you cannot reach. The same idea, used twice, settles the line across the middle of a trapezoid: it is parallel to both bases, and its length is their average.
Definition
A midsegment of a triangle is a segment whose two endpoints are the midpoints of two sides of the triangle. A triangle has three sides, so there are three ways to choose two of them, and every triangle has exactly three midsegments.
In the figure, D is the midpoint of \overline{AB} and E is the midpoint of \overline{AC}, so \overline{DE} is a midsegment. The side it does not touch, \overline{BC}, is called its third side. Matching tick marks show equal lengths: AD = DB and AE = EC.
A midsegment is not a median. A median has one end at a vertex and the other at a midpoint; a midsegment has both ends at midpoints and no end at a vertex.
The three midsegments together form a triangle inside the original, whose vertices are the three midpoints. It is called the medial triangle.
A trapezoid, as in trapezoids and kites, is a quadrilateral with at least one pair of parallel sides, called its bases. The other two sides are its legs. The midsegment of a trapezoid is the segment joining the midpoints of its two legs. Under this definition a parallelogram is a trapezoid too; it has two pairs of parallel sides, and either pair may be taken as the bases, with the other pair as the legs.
Intuition
Stand at vertex A and walk halfway along each of the two sides that leave it, to D and to E. The small triangle ADE is the whole triangle ABC shrunk by half toward A. Every distance measured from A along those sides is halved, so the far side \overline{DE} comes out half the length of \overline{BC} and pointing the same way. The shape of the triangle never enters, because the shrinking is the same in every direction.
For a trapezoid, the midsegment sits halfway between the two bases. A segment halfway between two parallel sides should have a length halfway between theirs, and halfway between b_1 and b_2 is their average, \tfrac12(b_1 + b_2). Two extreme cases agree. In a rectangle both bases are the same length and so is the midsegment. Squeeze the top base down to a point, of length 0, and the trapezoid becomes a triangle; the average is then \tfrac12(0 + b) = \tfrac12 b: exactly the triangle result.
Results
Triangle Midsegment Theorem. If D and E are the midpoints of sides \overline{AB} and \overline{AC} of \triangle ABC, then
Converse: the parallel through a midpoint. A line through the midpoint of one side of a triangle, parallel to a second side, meets the third side at its midpoint.
The medial triangle. The three midsegments cut a triangle with sides a, b and c into four congruent triangles, each with sides \tfrac12 a, \tfrac12 b and \tfrac12 c. So the medial triangle has half the perimeter of the original and a quarter of its area.
Trapezoid Midsegment Theorem. In trapezoid ABCD with bases \overline{AB} and \overline{DC}, let M and N be the midpoints of the legs \overline{AD} and \overline{BC}. Then
The midsegment of a trapezoid is parallel to both bases, and its length is the average of theirs.
Why it is true
The triangle theorem, through a parallelogram. Start with D and E the midpoints of \overline{AB} and \overline{AC}. Extend \overline{DE} past E to a point F chosen so that EF = DE, and join F to C.
First compare \triangle ADE with \triangle CFE. AE = CE because E is the midpoint of \overline{AC}. DE = FE because that is how F was chosen. \angle AED \cong \angle CEF because they are vertical angles. Two sides and the angle between them match, so the triangles are congruent by SAS (the congruence criteria).
Matching parts of congruent triangles are equal, so CF = AD and \angle DAE \cong \angle FCE. Those two angles sit on opposite sides of the line AC, between the lines AB and CF: they are alternate interior angles, and since they are equal, \overline{CF} \parallel \overline{AB} (alternate interior angles).
Now look at quadrilateral DBCF. AD = DB because D is a midpoint, so CF = DB as well. Its opposite sides \overline{DB} and \overline{FC} are therefore parallel and equal, and one pair of opposite sides parallel and equal is enough to make a quadrilateral a parallelogram (proving a quadrilateral is a parallelogram).
In a parallelogram the other pair of opposite sides is parallel and equal too, so \overline{DF} \parallel \overline{BC} and DF = BC. E is the midpoint of \overline{DF} by construction, so \overline{DE} lies along \overline{DF} and is half of it:
The same fact, seen as a dilation. Take the dilation with center A and scale factor 2. It sends D to B, because B lies on ray AD at twice the distance from A, and for the same reason it sends E to C. So it sends \overline{DE} to \overline{BC}. A dilation takes a segment that does not pass through its center to a parallel segment, with every length multiplied by the scale factor (what a dilation does to a line). So \overline{BC} \parallel \overline{DE} and BC = 2\,DE, which is the theorem again.
The converse. Let D be the midpoint of \overline{AB}. Through D there is only one line parallel to \overline{BC}. The midsegment from D to the midpoint E of \overline{AC} is parallel to \overline{BC}, by the theorem, so it lies on that one line. A line through D parallel to \overline{BC} is therefore the line DE, and it meets \overline{AC} at E, the midpoint.
The trapezoid theorem, from the triangle theorem twice. In trapezoid ABCD with \overline{AB} \parallel \overline{DC}, draw the diagonal \overline{AC} and call its midpoint P.
In \triangle ACD, M and P are the midpoints of \overline{AD} and \overline{AC}, so \overline{MP} is a midsegment: \overline{MP} \parallel \overline{DC} and MP = \tfrac12\, DC.
In \triangle CAB, P and N are the midpoints of \overline{CA} and \overline{CB}, so \overline{PN} is a midsegment: \overline{PN} \parallel \overline{AB} and PN = \tfrac12\, AB.
The bases are parallel, so the lines MP and PN both pass through P and are both parallel to \overline{DC}. Only one line through P is parallel to \overline{DC}, so they are the same line: M, P and N lie on one line, parallel to both bases. One more fact is taken from the picture rather than proved here: a trapezoid is convex, so its diagonals cross inside it and \overline{AC} separates D from B. Then M (on the leg through D) and N (on the leg through B) are on opposite sides of \overline{AC}, and P lies between them. The two pieces add:
Worked examples
A length and two angles from one midsegment
In \triangle ABC, D and E are the midpoints of \overline{AB} and \overline{AC}. You are given BC = 18 and m\angle ABC = 64^\circ. Find DE, m\angle ADE and m\angle BDE.
\overline{DE} joins the midpoints of two sides, so it is a midsegment, and its third side is \overline{BC}. By the Triangle Midsegment Theorem it is half as long:
The same theorem says \overline{DE} \parallel \overline{BC}. Read \overline{AB} as a transversal crossing these two parallels. \angle ADE and \angle ABC sit in matching positions at the two crossings, so they are corresponding angles and are equal: m\angle ADE = 64^\circ.
\angle BDE and \angle ADE together make the straight angle at D along \overline{AB}, so they form a linear pair and add to 180^\circ: m\angle BDE = 180^\circ - 64^\circ = 116^\circ.
So DE = 9, m\angle ADE = 64^\circ and m\angle BDE = 116^\circ.
A midsegment written in terms of x
In \triangle ABC, D and E are the midpoints of \overline{AB} and \overline{AC}, with DE = 2x + 1 and BC = 5x - 4. Find x, DE and BC.
SolutionThe third side is twice the midsegment, so BC = 2 \cdot DE. Writing it this way round avoids a fraction:
Then DE = 2(6) + 1 = 13 and BC = 5(6) - 4 = 26, so DE = 13 and BC = 26.
Check. 26 is twice 13, as the theorem demands.
The triangle of midpoints
A right triangle has legs of length 12 and 16 and a hypotenuse of length 20. Its three midpoints are joined to make the medial triangle. Find the perimeter and the area of the medial triangle.
Each side of the medial triangle is a midsegment, and each is half of the side it does not touch. So the medial triangle has sides 6, 8 and 10: the midsegment parallel to the leg of 12 is 6, the one parallel to the leg of 16 is 8, and the one parallel to the hypotenuse is 10. Its perimeter is 6 + 8 + 10, which gives perimeter 24, half of the original 12 + 16 + 20 = 48.
For the area, notice that 6^2 + 8^2 = 36 + 64 = 100 = 10^2, so by the converse of the Pythagorean theorem the medial triangle is right-angled with legs 6 and 8. Its area is \tfrac12 \cdot 6 \cdot 8, which gives area 24.
Check by the four pieces. The original area is \tfrac12 \cdot 12 \cdot 16 = 96. Each corner triangle has two half-sides and one midsegment, which are again 6, 8 and 10, so all four small triangles are congruent by SSS and each is a quarter of 96, which is 24. The two methods agree.
Testing the theorem with coordinates
\triangle ABC has vertices A(2, 8), B(-4, 0) and C(10, 2). D and E are the midpoints of \overline{AB} and \overline{AC}. Show that \overline{DE} \parallel \overline{BC} and that DE = \tfrac12\, BC.
SolutionA midpoint’s coordinates are the averages of the endpoints’ coordinates:
So D = (-1, 4) and E = (6, 5). For direction, compare slopes:
Equal slopes on two different lines mean the lines are parallel, so \overline{DE} \parallel \overline{BC}. For length, use the distance formula:
\displaystyle 5\sqrt2 is half of \displaystyle 10\sqrt2, so DE = \tfrac12\, BC.
This checks one triangle, not every triangle. Placing a general triangle at coordinates with letters in place of numbers turns the same calculation into a proof, which is what coordinate proofs with variables do.
Using the converse
In \triangle ABC, D is the midpoint of \overline{AB}. The line through D parallel to \overline{BC} meets \overline{AC} at E. You are given AE = 3x - 1, EC = x + 7 and BC = 30. Find x, AC and DE.
SolutionNothing says that E is a midpoint, so the theorem cannot be used yet. The converse supplies it: the line through the midpoint D, parallel to \overline{BC}, meets the third side at its midpoint. So AE = EC:
Then AE = 3(4) - 1 = 11 and EC = 4 + 7 = 11, so AC = 22.
Now D and E are both midpoints, so \overline{DE} is a midsegment with third side \overline{BC}, and DE = \tfrac12 \cdot 30, so DE = 15.
The midsegment of a trapezoid
(a) Trapezoid ABCD has bases AB = 14 and DC = 22, and M and N are the midpoints of the legs. Find MN.
(b) Another trapezoid has one base of 11 and a midsegment 15 long. Find the other base.
(a) The midsegment of a trapezoid is the average of the bases:
so MN = 18. It lies between 14 and 22, as a line halfway between the bases should.
(b) Call the other base b. The midsegment is the average of the bases, so
Check. \tfrac12(11 + 19) = \tfrac12(30) = 15.
A trapezoid in terms of x
A trapezoid has bases of lengths x + 4 and 3x - 2, and its midsegment has length 3x - 5. Find x and all three lengths.
SolutionThe midsegment is the average of the bases:
The bases are 6 + 4 = 10 and 3(6) - 2 = 16, so the bases are 10 and 16, and the midsegment is 3(6) - 5 = 13.
Check. \tfrac12(10 + 16) = 13, and every length is positive, so the trapezoid can exist.
The midpoints of any quadrilateral
ABCD is any quadrilateral whose sides do not cross each other, and P, Q, R and S are the midpoints of \overline{AB}, \overline{BC}, \overline{CD} and \overline{DA}. Show that PQRS is a parallelogram, and that its perimeter equals the sum of the diagonals of ABCD.
There is no triangle in the problem yet, so make one. The diagonal \overline{AC} gives two triangles, \triangle ABC and \triangle ADC. (If ABCD caves in, this diagonal may run outside it; the two triangles are still there, and nothing below needs them inside.)
In \triangle ABC, P and Q are the midpoints of \overline{AB} and \overline{BC}, so \overline{PQ} is a midsegment with third side \overline{AC}: \overline{PQ} \parallel \overline{AC} and PQ = \tfrac12\, AC.
In \triangle ADC, S and R are the midpoints of \overline{AD} and \overline{DC}, so \overline{SR} is a midsegment with the same third side: \overline{SR} \parallel \overline{AC} and SR = \tfrac12\, AC.
Both are parallel to \overline{AC}, so they are parallel to each other, and both are \tfrac12\, AC long. One pair of opposite sides of PQRS is parallel and equal, so PQRS is a parallelogram.
The other diagonal does the same for the other pair: \triangle BCD and \triangle BAD give QR = \tfrac12\, BD and PS = \tfrac12\, BD. Adding all four sides,
and the right-hand side collects to AC + BD, so the perimeter of PQRS is AC + BD.
Nothing in the argument needed ABCD to have any parallel sides or any particular shape. The result is usually called Varignon’s theorem, and it is the midsegment theorem used twice for something it was never stated for.
Boundaries
Both ends must be midpoints. A segment joining points one third of the way along two sides from the same vertex is still parallel to the third side, but it is one third as long, not half. That general case, a line parallel to one side cutting the other two in any ratio, is the triangle proportionality theorem.
Parallel alone does not make a midsegment. A segment across a triangle parallel to one side can have any length less than that side. It is half as long only when it starts at a midpoint, which is exactly what the converse adds.
A trapezoid’s midsegment joins the legs, not the bases. The segment joining the midpoints of the two bases follows no such rule: its length and direction change as the trapezoid leans.
Without a pair of parallel sides, the average fails. Join the midpoints of two opposite sides of a quadrilateral. Unless the other two sides are parallel, the segment is strictly shorter than half their sum (Exercise 25). Equality is exactly the trapezoid case, with those two sides as the bases.
A parallelogram works either way. Either pair of opposite sides can be the bases. The midsegment is then the average of two equal sides, so it equals each of them.
The figure must be a real triangle. If A, B and C lie on one line there is no triangle, and “parallel to the third side” has nothing to say: the midsegment lies along the same line.
Where this goes next
The small triangle ADE cut off by a midsegment has the same angles as \triangle ABC and sides half as long, which makes it the first example of similar triangles. Replace “half” by any fraction and you have the triangle proportionality theorem, which says a line parallel to one side splits the other two sides in the same ratio.
The midsegment also explains a fact about medians. If the medians from B and C meet at G, the midsegment joining their feet is parallel to \overline{BC} and half as long, and that forces G to be two thirds of the way along each median: the centroid.
The trapezoid result gives the area of a trapezoid in one line, as the midsegment times the height, which is the average of the bases times the height (area of parallelograms and trapezoids). And the coordinate check in Example 4 becomes a proof for every triangle once the vertices are written with letters, which is the starting point of coordinate proofs with variables.
Exercises
Triangle midsegments
In Exercises 1 to 3, D and E are the midpoints of \overline{AB} and \overline{AC} in \triangle ABC.
- If BC = 26, find DE.
- If DE = 7.5, find BC.
- If m\angle ACB = 48^\circ, find m\angle AED and m\angle DEC.
- A triangle has sides 8, 11 and 15. Find the perimeter of its medial triangle.
- The medial triangle of a triangle has sides 5, 6 and 9. Find the perimeter of the original triangle.
- A triangle has area 60. Find the area of its medial triangle.
Using algebra
In Exercises 7 to 9, D and E are the midpoints of \overline{AB} and \overline{AC} in \triangle ABC.
- DE = x + 5 and BC = 3x - 4. Find x, DE and BC.
- DE = 4x - 3 and BC = 5x + 9. Find x, DE and BC.
- AD = 2y + 1, DB = 3y - 4 and DE = y + 4. Find y, AB and BC.
- In \triangle ABC, D is the midpoint of \overline{AB}, and the line through D parallel to \overline{BC} meets \overline{AC} at E. If AE = 5x - 3 and EC = 2x + 9, find x and AC.
Coordinates
- \triangle ABC has vertices A(0, 0), B(8, 2) and C(4, 10). Find the midpoints D of \overline{AC} and E of \overline{BC}, and show that \overline{DE} \parallel \overline{AB} and DE = \tfrac12\, AB.
- The midpoints of the three sides of a triangle are (1, 3), (4, 5) and (2, -1). Find the vertices of the triangle.
Trapezoids
- A trapezoid has bases 9 and 17. Find the length of its midsegment.
- A trapezoid has a midsegment of length 21 and one base of length 26. Find the other base.
- A trapezoid has bases 2x + 1 and 4x - 3 and a midsegment of length 17. Find x and both bases.
- Trapezoid ABCD has bases AB = 20 and DC = 12. M and N are the midpoints of the legs \overline{AD} and \overline{BC}, and \overline{MN} crosses the diagonal \overline{AC} at P. Find MP, PN and MN.
- A parallelogram has sides 10 and 6. Find the length of the segment joining the midpoints of the two sides of length 6.
- A garden bed is a trapezoid whose parallel edges are 3.6 m and 5.2 m long. A straight path joins the midpoints of the other two edges. How long is the path?
Repeating the construction
- Start with a triangle of perimeter 48 and area 128. Draw its medial triangle, then the medial triangle of that, then the medial triangle of that one. Find the perimeter and the area of the last triangle drawn.
- A quadrilateral has diagonals of lengths 10 and 14. Find the perimeter of the quadrilateral made by joining the midpoints of its sides in order.
Reasoning
- Explain why the three midsegments of a triangle cut it into four congruent triangles.
- Prove the converse of the Triangle Midsegment Theorem: a line through the midpoint of one side of a triangle, parallel to a second side, meets the third side at its midpoint.
- Show that the midsegment of a trapezoid passes through the midpoints of both diagonals.
- Show that joining the midpoints of the sides of a rectangle, in order, gives a rhombus.
- ABCD is a quadrilateral with no two sides parallel, and M and N are the midpoints of \overline{AD} and \overline{BC}. Show that MN is less than \tfrac12(AB + DC).
Answers
- DE = 13. A midsegment is half its third side: \tfrac12 \cdot 26.
- BC = 15. The third side is twice the midsegment: 2 \cdot 7.5.
- m\angle AED = 48^\circ and m\angle DEC = 132^\circ. \overline{DE} \parallel \overline{BC} with transversal \overline{AC}, so \angle AED and \angle ACB are corresponding angles; \angle DEC forms a linear pair with \angle AED, so it is 180^\circ - 48^\circ.
- 17. The medial triangle has sides 4, 5.5 and 7.5, half of each side, and half of 8 + 11 + 15 = 34 is 17.
- 40. Each side of the original is twice a side of the medial triangle: 10, 12 and 18.
- 15. The midsegments cut the triangle into four congruent triangles, one of which is the medial triangle, so it is \tfrac14 \cdot 60.
- x = 14, DE = 19 and BC = 38. From BC = 2 \cdot DE: 3x - 4 = 2x + 10.
- x = 5, DE = 17 and BC = 34. From 5x + 9 = 2(4x - 3) = 8x - 6, 3x = 15.
- y = 5, AB = 22 and BC = 18. D is the midpoint, so 2y + 1 = 3y - 4; then AD = DB = 11, and DE = 9 is half of BC.
- x = 4 and AC = 34. By the converse, E is the midpoint of \overline{AC}, so 5x - 3 = 2x + 9; then AE = EC = 17.
- D = (2, 5) and E = (6, 6). The slope of \overline{DE} is \tfrac{6 - 5}{6 - 2} = \tfrac14 and the slope of \overline{AB} is \tfrac{2}{8} = \tfrac14, so they are parallel. \displaystyle DE = \sqrt{4^2 + 1^2} = \sqrt{17} and \displaystyle AB = \sqrt{8^2 + 2^2} = \sqrt{68} = 2\sqrt{17}, so DE = \tfrac12\, AB.
- (3, 9), (-1, -3) and (5, 1). If the vertices are X, Y, Z with midpoints \tfrac12(X + Y) = (1, 3), \tfrac12(Y + Z) = (4, 5) and \tfrac12(Z + X) = (2, -1), adding gives X + Y + Z = (7, 7), and each vertex is (7, 7) minus twice the midpoint of the side opposite it. Check: the midpoint of (3, 9) and (-1, -3) is (1, 3), and so on.
- 13. The average of the bases: \tfrac12(9 + 17).
- 16. From \tfrac12(26 + b) = 21, 26 + b = 42.
- x = 6, and the bases are 13 and 21. The average of the bases is \tfrac12(6x - 2) = 3x - 1, so 3x - 1 = 17.
- MP = 6, PN = 10 and MN = 16. \overline{MN} is parallel to \overline{DC} through the midpoint M of \overline{AD}, so by the converse in \triangle ACD it meets \overline{AC} at its midpoint; then MP = \tfrac12 \cdot 12 in \triangle ACD and PN = \tfrac12 \cdot 20 in \triangle ABC. Check: \tfrac12(20 + 12) = 16.
- 10. Take the sides of length 10 as the bases and the sides of length 6 as the legs; the midsegment is \tfrac12(10 + 10).
- 4.4 m. The path is the midsegment: \tfrac12(3.6 + 5.2) = \tfrac12(8.8).
- Perimeter 6 and area 2. Each step halves the perimeter and quarters the area, so three steps give 48 \div 8 and 128 \div 64.
- 24. By Example 8 the perimeter equals the sum of the diagonals, 10 + 14.
- Call the sides a, b, c. Each corner triangle has two half-sides and the midsegment facing the corner, for example \tfrac12 c, \tfrac12 b and \tfrac12 a at vertex A. The middle triangle’s sides are the three midsegments, \tfrac12 a, \tfrac12 b and \tfrac12 c. All four have the same three side lengths, so they are congruent by SSS.
- Let D be the midpoint of \overline{AB} and \ell the line through D parallel to \overline{BC}. Let E be the midpoint of \overline{AC}. By the theorem, \overline{DE} \parallel \overline{BC}. Through D there is only one parallel to \overline{BC}, so line DE is \ell, and \ell meets \overline{AC} at E, its midpoint.
- Let M and N be the midpoints of the legs \overline{AD} and \overline{BC} of trapezoid ABCD with bases \overline{AB} and \overline{DC}. \overline{MN} is parallel to the bases. In \triangle ACD, line MN passes through the midpoint M of \overline{AD} parallel to \overline{DC}, so by the converse it meets \overline{AC} at its midpoint. In \triangle ABD, line MN passes through M parallel to \overline{AB}, so it meets \overline{BD} at its midpoint.
- By Example 8 the midpoint quadrilateral is a parallelogram whose sides are half the diagonals: two sides are \tfrac12 of one diagonal and two are \tfrac12 of the other. A rectangle’s diagonals are equal, so all four sides are equal, and a quadrilateral with four equal sides is a rhombus.
- Let P be the midpoint of \overline{AC}. In \triangle ACD, MP = \tfrac12\, DC and \overline{MP} \parallel \overline{DC}; in \triangle ABC, PN = \tfrac12\, AB and \overline{PN} \parallel \overline{AB}. By the triangle inequality, MN \le MP + PN = \tfrac12(AB + DC), with equality only if P lies on \overline{MN}. Then \overline{MP} and \overline{PN} lie on one line, parallel to both \overline{DC} and \overline{AB}, which would make \overline{AB} \parallel \overline{DC}. No two sides are parallel, so the inequality is strict.