The Diagonals of a Parallelogram Bisect Each Other
?How to use▶
▶ 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
Drag A, B or D; the fourth vertex C and the crossing point M follow, and the readout under the canvas keeps M at the midpoint of both diagonals. Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?The diagonals AC and BD meet at M. Opposite sides are equal, AB = CD, and each diagonal is a transversal of the parallels AB, DC, so ∠BAC = ∠DCA and ∠ABD = ∠CDB. Triangles ABM and CDM then have the equal sides AB, CD between two equal angles: congruent (ASA), so AM = MC and BM = MD. Drag A, B or D; C and M follow.Parallelogram: diagonals
Display
Drag A, B or D; C and M follow
AM = 2.96, MC = 2.96; BM = 2.18, MD = 2.18
Step 1 / 6
Step explanations
1A parallelogram and its diagonals
Given ABCD is a parallelogram: AB ∥ DC and AD ∥ BC. Its diagonals AC and BD meet at M. To prove: AM = MC and BM = MD, the diagonals bisect each other.
Step through the proof, then drag the vertices and watch the crossing point of the diagonals stay at the midpoint of both. 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?Parallelogram: diagonalsDisplayLabelsMeasuresDrag A, B or D; C and M follow
AM = 2.96, MC = 2.96; BM = 2.18, MD = 2.18
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1A parallelogram and its diagonals
Given ABCD is a parallelogram: AB ∥ DC and AD ∥ BC. Its diagonals AC and BD meet at M. To prove: AM = MC and BM = MD, the diagonals bisect each other.
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1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc
3Draggable points — A, B and D move freely; the hollow C keeps the figure a parallelogram and the hollow M is where the diagonals cross
4Readout — the line under the figure: the four half-diagonals 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?Parallelogram: diagonalsDisplayLabelsMeasuresDrag A, B or D; C and M follow
AM = 2.96, MC = 2.96; BM = 2.18, MD = 2.18
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1A parallelogram and its diagonals
Given ABCD is a parallelogram: AB ∥ DC and AD ∥ BC. Its diagonals AC and BD meet at M. To prove: AM = MC and BM = MD, the diagonals bisect each other.
2AB = CD
Opposite sides Opposite sides of a parallelogram are equal, so AB = CD. These two sides belong to the triangles ABM and CDM, the two triangles that the diagonals cut off at the top and bottom.
3∠BAC = ∠DCA = α
Alternate angles AB ∥ DC and the diagonal AC crosses both. ∠BAC at A and ∠DCA at C are alternate angles, so they are equal.
4∠ABD = ∠CDB = β
Alternate angles AB ∥ DC and the diagonal BD crosses both. ∠ABD at B and ∠CDB at D are alternate angles too, so they are equal.
5Triangles ABM and CDM are congruent
ASA congruence In triangle ABM the side AB lies between α at A and β at B; in triangle CDM the equal side CD lies between α at C and β at D. Two angles and the included side: the triangles are congruent, with A ↔ C, B ↔ D, M ↔ M.
6AM = MC, BM = MD
Corresponding parts Corresponding sides of congruent triangles are equal: AM ↔ CM and BM ↔ DM. So M is the midpoint of AC and the midpoint of BD: each diagonal cuts the other in half. ∎
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1The crossing point M — the midpoint of both diagonals: one tick on each half of AC, two on each half of BD
2The two triangles — ABM below and CDM above, congruent by ASA on the equal bases AB and CD
3Alternate angles — α at A and C on the diagonal AC, β at B and D on the diagonal BD
5Replay — at the last step Play becomes Replay and Next is off
The Statement
The two diagonals of a parallelogram ABCD are AC and BD. They cross at a point M inside the figure, and the theorem says that M is the midpoint of both:
AM=MC,BM=MD
the diagonals bisect each other. It is the second basic property of a parallelogram, after the equal opposite sides, and the two are proved the same way, with a pair of congruent triangles made by alternate angles on a diagonal.
In the player C and M are hollow: C is computed as the fourth vertex that makes ABCD a parallelogram, and M as the crossing point of the diagonals, so both follow when A, B or D is dragged. The readout under the canvas shows the four half-diagonals in the frame's own units, live while you drag.
Opening scene, frozen
ABCD with its diagonals AC and BD crossing at the hollow point M.
Opposite Sides Equal
The diagonals cut the parallelogram into four triangles. The proof uses the two that sit on the parallel sides AB and DC: triangle ABM at the bottom and triangle CDM at the top, filled in two colours. They touch only at M, and they look like reflections of each other through that point; the proof shows that they are congruent.
The first fact needed is a side: AB=CD, because opposite sides of a parallelogram are equal, which is the previous page's theorem. Those two sides are the bases of the two triangles, and the congruence will be built on them.
Opposite sides equal, frozen
AB = CD lit; the two triangles ABM and CDM on those sides.
Alternate Angles Once
The diagonal AC crosses the parallel sides AB and DC: it is a transversal of them. The angle ∠BAC at A and the angle ∠DCA at C lie on opposite sides of the transversal, between the parallels, so they are alternate angles, and by the parallel lines theorem they are equal:
∠BAC=∠DCA=α
In the triangles, α is the angle of ABM at A and the angle of CDM at C; both sit at one end of the base.
Alternate angles once, frozen
On the transversal AC: ∠BAC = ∠DCA = α.
Alternate Angles Twice
The other diagonal, BD, is a transversal of the same two parallels AB and DC. The angles ∠ABD at B and ∠CDB at D are alternate angles on it, so they are equal as well:
∠ABD=∠CDB=β
Now β is the angle of triangle ABM at B and the angle of triangle CDM at D, at the other end of each base. Only the one pair of parallels, AB∥DC, has been used; the second pair, AD∥BC, was needed on the previous page to get AB=CD, and it enters this proof only through that.
Alternate angles twice, frozen
On the transversal BD: ∠ABD = ∠CDB = β.
Two Congruent Triangles
Put the three facts together. In triangle ABM the base AB lies between the angle α at A and the angle β at B. In triangle CDM the base CD lies between the angle α at C and the angle β at D. The bases are equal and so are the angles at their ends, so the triangles are congruent by angle-side-angle, with the vertices matched A↔C, B↔D, M↔M.
As on the previous page, angle-side-angle works because two angles fix the third by the angle sum, and a triangle with a given side and the two angles at its ends is determined. The matching of vertices is what makes the next step mechanical.
The congruent triangles, frozen
Equal bases between α and β: congruent by ASA.
Why It Holds
Corresponding sides of congruent triangles are equal. With A↔C and M↔M, the side AM corresponds to CM, so AM=MC; with B↔D, the side BM corresponds to DM, so BM=MD. The point M is the midpoint of AC and the midpoint of BD at once. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given parallelogram, equal opposite sides, alternate angles twice, angle-side-angle congruence, and corresponding parts.
A slightly different route uses only one pair of alternate angles together with the vertical angles at M: ∠AMB=∠CMD, since they are opposite angles at the crossing, and with α and AB=CD that is angle-angle-side. The picture is the same; the choice is only which equal angles to point at.
The conclusion, frozen
AM = MC (one tick each) and BM = MD (two ticks each).
Dragging the Figure
Drag D sideways: the parallelogram shears, the diagonals change length and tilt, M slides with them, and the readout keeps AM=MC and BM=MD to the last digit. Drag D up or down to change the height, or A and B to change the base; C and M follow in every case.
Notice that the two diagonals are usually of different lengths, and that they usually cross at an angle other than 90°. Both of those are properties of special parallelograms, equal diagonals of a rectangle and perpendicular diagonals of a rhombus, and both have their own pages; bisecting each other is what every parallelogram does. A drag that would flatten the figure, or carry a vertex out of the frame, is refused.
After dragging D, frozen
D pushed right: the diagonals are now very different in length, and M still halves both.
Consequences
• The converse. If the diagonals of a quadrilateral bisect each other, it is a parallelogram: the triangles ABM and CDM are then congruent by side-angle-side, using the vertical angles at M, and the equal alternate angles make AB∥DC. • A centre of symmetry. Reflecting the parallelogram through M sends A to C and B to D, so the figure maps onto itself; M is its centre. • Rectangles and rhombi. A rectangle's diagonals are also equal, and a rhombus's are also perpendicular; a square has both. Each is the parallelogram property plus one more. • Vectors and coordinates. The midpoint of AC is 21(A+C) and the midpoint of BD is 21(B+D); the theorem says these are the same point, which is the coordinate form of A+C=B+D, the parallelogram rule. • Locating the centre. The centre of a parallelogram-shaped region, its balance point, is where the diagonals cross.