▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Slide C along the top base, or drag A, B or D; the midpoints follow, and the readout under the canvas keeps MN at half the sum of the bases. Learn more about why it holds
Proof?Draw the diagonal AC and mark its midpoint K. In triangle ACD, MK is a midsegment, so MK ∥ DC and MK = ½ DC; in triangle ABC, KN is a midsegment, so KN ∥ AB and KN = ½ AB. Both are parallel to the bases and pass through K, so they lie on one line: M, K, N are collinear and MN = ½ (AB + DC). Drag C along the top, or any other vertex.Trapezoid midsegment
Display
Drag C along the top base, A, B or D to reshape the trapezoid
MN = 4.39, (AB + DC) / 2 = 4.39
Step 1 / 6
Step explanations
1The midpoints of the legs
Given ABCD is a trapezoid with AB ∥ DC (one chevron each). M is the midpoint of the leg AD and N the midpoint of the leg BC. To prove: MN ∥ AB and MN = ½ (AB + DC).
Step through the proof, then reshape the trapezoid and watch the segment between the midpoints of its legs stay parallel to the bases, always half their sum. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
Proof?Trapezoid midsegmentDisplayLabelsMeasuresDrag C along the top base, A, B or D to reshape the trapezoid
MN = 4.39, (AB + DC) / 2 = 4.39
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1The midpoints of the legs
Given ABCD is a trapezoid with AB ∥ DC (one chevron each). M is the midpoint of the leg AD and N the midpoint of the leg BC. To prove: MN ∥ AB and MN = ½ (AB + DC).
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1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names
3Draggable points — C slides along the top base, A, B and D move freely; the hollow midpoints M, N and later K follow
4Readout — the line under the figure: MN and half the sum of the bases, in the frame's units, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
Proof?Trapezoid midsegmentDisplayLabelsMeasuresDrag C along the top base, A, B or D to reshape the trapezoid
MN = 4.39, (AB + DC) / 2 = 4.39
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1The midpoints of the legs
Given ABCD is a trapezoid with AB ∥ DC (one chevron each). M is the midpoint of the leg AD and N the midpoint of the leg BC. To prove: MN ∥ AB and MN = ½ (AB + DC).
2The diagonal AC and its midpoint K
Construction Draw the diagonal AC and mark its midpoint K (three ticks each). The diagonal cuts the trapezoid into two triangles, ACD and ABC, and K is a midpoint in both.
3MK ∥ DC and MK = ½ DC
Midsegment theorem In triangle ACD, M is the midpoint of AD and K the midpoint of AC, so MK is a midsegment: it is parallel to the third side DC and half as long.
4KN ∥ AB and KN = ½ AB
Midsegment theorem In triangle ABC, K is the midpoint of AC and N the midpoint of BC, so KN is a midsegment: it is parallel to the third side AB and half as long.
5M, K and N lie on one line
One parallel through a point DC ∥ AB, so MK and KN are both parallel to AB, and both pass through K. Through a point there is only one line parallel to a given line, so MK and KN are parts of the same line: K lies on MN, and MN ∥ AB.
6MN = ½ (AB + DC)
Adding the parts Since K is between M and N, MN = MK + KN = ½ DC + ½ AB = ½ (AB + DC): the midsegment is parallel to the bases (one chevron) and equal to half their sum. ∎
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1The midsegment MN — through the midpoint K of the diagonal, parallel to the bases, half their sum
2The two triangles — ACD above the diagonal with MK, ABC below it with KN
3The diagonal AC — dashed, with its midpoint K marked by three ticks on each half
4Justification tags — Given, Construction, Midsegment theorem (twice), One parallel through a point, Adding the parts
5Replay — at the last step Play becomes Replay and Next is off
The Statement
A trapezoid is a quadrilateral with one pair of parallel sides, the bases; the other two sides are the legs. In ABCD the bases are AB and DC, the legs AD and BC, and M, N are the midpoints of the legs. The segment MN is the midsegment (or median) of the trapezoid, and the theorem says
MN∥AB,MN=21(AB+DC)
it is parallel to the bases and equal to their average. The midsegment theorem for triangles is the case DC=0, when the top base shrinks to a point; this page's proof reduces the trapezoid to two triangles and uses that theorem in each.
In the player C slides along the line through D parallel to AB, which is what keeps the figure a trapezoid; A, B and D move freely, and the midpoints are computed. The readout under the canvas shows MN and half the sum of the bases, in the frame's own units, live while you drag.
Opening scene, frozen
The trapezoid with AB ∥ DC, the midpoints M and N of the legs, and the midsegment MN.
The Diagonal and Its Midpoint
The construction is the diagonal AC, dashed in the player, together with its midpoint K, marked with three ticks on each half. The diagonal cuts the trapezoid into two triangles: ACD, which has the top base DC as a side, and ABC, which has the bottom base AB as a side.
The point K is the hinge of the proof. It is a midpoint in both triangles at once, so it pairs with M in the first and with N in the second, and each pair is a midsegment of its triangle. Everything else is the midsegment theorem and one fact about parallel lines.
The diagonal and its midpoint, frozen
AC dashed, K its midpoint with three ticks each side.
The First Midsegment
In triangle ACD, M is the midpoint of AD and K the midpoint of AC, so MK joins the midpoints of two sides. By the midsegment theorem,
MK∥DC,MK=21DC
The player fills the triangle and draws MK in the same colour. It is the part of the eventual midsegment that belongs to the top base.
The first midsegment, frozen
MK in triangle ACD: parallel to DC and half of it.
The Second Midsegment
In triangle ABC, K is the midpoint of AC and N the midpoint of BC, so KN is a midsegment of that triangle, and the same theorem gives
KN∥AB,KN=21AB
This is the part that belongs to the bottom base. So far MK and KN are two separate segments meeting at K; nothing yet says that they make a straight line, which is what the next step is about.
The second midsegment, frozen
KN in triangle ABC: parallel to AB and half of it.
One Line Through K
The bases are parallel, DC∥AB, so MK, which is parallel to DC, is also parallel to AB. Now both MK and KN are lines through K parallel to AB. Through a point not on a line there is exactly one line parallel to it; that is the parallel postulate, the fact behind the whole parallel lines page. So MK and KN are pieces of one and the same line.
In other words M, K and N are collinear, with K between M and N, and the line through them is parallel to AB. That settles the first half of the theorem: MN∥AB, and MN passes through the midpoint of the diagonal.
One line through K, frozen
MK and KN both parallel to the bases through K: one line.
Why It Holds
Since K lies between M and N, the length MN is the sum of the two pieces:
MN=MK+KN=21DC+21AB=21(AB+DC)
The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given trapezoid, a construction, the midsegment theorem twice, the uniqueness of the parallel through a point, and an addition.
The same diagonal shows a second fact. The other diagonal BD has a midpoint too, and by the same argument it lies on MN as well; the segment between the two diagonal midpoints has length 21(AB−DC), half the difference of the bases, and it is the part of MN that lies between the two diagonals.
The conclusion, frozen
MN = MK + KN = half the sum of the bases, with its chevron.
Dragging the Figure
Slide C along the top: the top base grows or shrinks, N and K move with it, and the readout keeps MN at exactly half the sum of the bases. Push C all the way towards D and the trapezoid nearly becomes a triangle, with MN nearly half of AB, which is the triangle midsegment theorem appearing as a limit. Drag D to change the height or shear the figure, A or B to change the bottom base.
C cannot leave the line through D parallel to AB, since then the figure would not be a trapezoid. A drag that would flatten the trapezoid, shrink the top base to nothing, or carry a vertex out of the frame, is refused.
After dragging C, frozen
C slid towards D: a short top base, MN not much more than half of AB, still the average.
Consequences
• Area. The area of a trapezoid is its height times the midsegment, h⋅21(AB+DC): cut along MN and turn the top piece over to make a parallelogram with base MN and the same height, whose area is base times height by the parallelogram pages. • The triangle as a limit. With DC=0 the trapezoid is a triangle and the theorem becomes the triangle midsegment theorem. • The diagonal midpoints. Both diagonal midpoints lie on MN, and the segment between them is half the difference of the bases. • Varignon again. In an isosceles trapezoid the diagonals are equal, so by Varignon's theorem the midpoints of its four sides form a rhombus; two of those midpoints are M and N. • Dividing a segment. Equally spaced parallel lines cut every transversal into equal parts; the midsegment is the case of three lines, and it is how a segment is divided into equal parts with a ruler and set square.