▶ 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 D round A to open or close the rhombus, or A and B to move it; C and M follow, and the readout under the canvas keeps both angles at M at 90°. 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?All four sides of a rhombus are equal, so it is a parallelogram and its diagonals bisect each other: BM = MD. Triangles ABM and ADM have AB = AD, BM = MD and the common side AM, so they are congruent (SSS) and the angles at M are equal. The two angles make a straight angle along BD, so each is 90°. Drag D round A to open or close the rhombus.Rhombus: diagonals
Display
Drag D round A to open or close the rhombus, A or B to move it
∠AMB = 90.0°, ∠AMD = 90.0°
Step 1 / 6
Step explanations
1A rhombus and its diagonals
Given ABCD is a rhombus: all four sides are equal (one tick each). Its diagonals AC and BD meet at M. To prove: AC ⟂ BD, the diagonals are perpendicular.
Step through the proof, then open and close the rhombus and watch its diagonals keep crossing at right angles. 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?Rhombus: diagonalsDisplayLabelsMeasuresDrag D round A to open or close the rhombus, A or B to move it
∠AMB = 90.0°, ∠AMD = 90.0°
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1A rhombus and its diagonals
Given ABCD is a rhombus: all four sides are equal (one tick each). Its diagonals AC and BD meet at M. To prove: AC ⟂ BD, the diagonals are perpendicular.
1234567
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 — D swings round A on a circle through B, which keeps the four sides equal; A and B move the figure; the hollow C and M follow
4Readout — the line under the figure: the two angles at M, 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?Rhombus: diagonalsDisplayLabelsMeasuresDrag D round A to open or close the rhombus, A or B to move it
∠AMB = 90.0°, ∠AMD = 90.0°
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1A rhombus and its diagonals
Given ABCD is a rhombus: all four sides are equal (one tick each). Its diagonals AC and BD meet at M. To prove: AC ⟂ BD, the diagonals are perpendicular.
2BM = MD
Parallelogram diagonals A rhombus has equal opposite sides, so it is a parallelogram, and the diagonals of a parallelogram bisect each other. So M is the midpoint of BD: BM = MD (two ticks each).
3Triangles ABM and ADM are congruent
SSS congruence AB = AD, sides of the rhombus; BM = MD, just shown; and AM is a side of both. Three pairs of equal sides: the triangles are congruent, with A ↔ A, B ↔ D, M ↔ M.
4∠AMB = ∠AMD = φ
Corresponding parts Corresponding angles of congruent triangles are equal. The angle at M in triangle ABM corresponds to the angle at M in triangle ADM, so the two angles at M on either side of AM are equal.
5φ + φ = 180°
Linear pair B, M and D lie on one line, the diagonal BD. The two angles ∠AMB and ∠AMD together fill the straight angle at M along that line, so they add up to 180°.
6AC ⟂ BD
Conclusion 2φ = 180°, so φ = 90°: the diagonals meet at right angles. The same congruence shows that AM bisects the angle at A, so each diagonal of a rhombus also bisects the angles at its ends. ∎
12345
1The right angle at M — the mark that appears at the last step: the diagonals are perpendicular
2The two triangles — ABM and ADM, congruent by SSS; their angles at M are the linear pair
3The half-diagonals — BM = MD, two ticks each, from the parallelogram theorem
5Replay — at the last step Play becomes Replay and Next is off
The Statement
A rhombus is a quadrilateral with four equal sides. Its two diagonals AC and BD cross at a point M, and the theorem says they cross at right angles:
AC⊥BD
A second fact comes out of the same proof: each diagonal bisects the two angles of the rhombus at its ends, so AM splits the angle at A in half, and the four small triangles around M are all congruent right triangles.
The proof builds on the two parallelogram pages. A rhombus is a parallelogram, so its diagonals bisect each other; that gives three equal pairs of sides in two triangles, and the rest is a straight angle cut into two equal parts. In the player D slides on a circle about A through B, which is what keeps AD=AB; C and M are computed. The readout under the canvas shows the two angles at M in degrees, live while you drag.
Opening scene, frozen
ABCD with four equal sides (one tick each) and its diagonals crossing at the hollow point M.
A Rhombus Is a Parallelogram
A rhombus has four equal sides, so in particular its opposite sides are equal, and a quadrilateral with both pairs of opposite sides equal is a parallelogram; that is the converse stated on the opposite sides page. So everything true of parallelograms is true of the rhombus, and the fact needed here is that the diagonals bisect each other:
BM=MD
The player marks the two halves of BD with two ticks each. Only this half of the parallelogram theorem is used; AM=MC is true as well but plays no part.
The diagonals bisect each other, frozen
BM = MD, two ticks each: the parallelogram theorem applied to the rhombus.
Three Equal Sides
Look at the two triangles on either side of the half-diagonal AM: triangle ABM and triangle ADM, filled in two colours. Compare their sides. AB=AD, because they are sides of the rhombus. BM=MD, from the previous step. And AM is a side of both. Three pairs of equal sides, so the triangles are congruent by side-side-side, with A↔A, B↔D, M↔M.
Side-side-side is the congruence rule used on the isosceles base angles page; it says a triangle is determined by its three sides. The picture here is in fact that page's picture: ABD is isosceles with apex A, and AM joins the apex to the midpoint of the base.
The congruent triangles, frozen
ABM and ADM: AB = AD, BM = MD, AM shared.
Equal Angles at M
Corresponding parts of congruent triangles are equal, and with M↔M the angle of triangle ABM at M corresponds to the angle of triangle ADM at M. So the two angles that AM makes with the diagonal BD, one on the side of B and one on the side of D, are equal; the player calls both φ.
The same congruence matches the angle at A in one triangle with the angle at A in the other, so AM bisects the angle ∠BAD. That is the second fact promised in The Statement, and it comes for free.
Equal angles at M, frozen
The two angles φ on either side of AM.
A Linear Pair
The two equal angles φ sit on opposite sides of AM along one straight line, the diagonal BD: B, M and D are collinear. Two angles that fill a straight angle in this way are a linear pair, and they are supplementary: they add up to 180°, as on the vertical angles page, where the same fact started the proof. So
φ+φ=180°
Nothing about the rhombus is used in this step; it is true of any two angles on a straight line that share a ray.
The linear pair, frozen
B, M, D on one line: the two angles fill a straight angle.
Why It Holds
2φ=180° gives φ=90°: the angle between the diagonals is a right angle, so AC⊥BD. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given rhombus, the parallelogram diagonals theorem, side-side-side congruence, corresponding parts, a linear pair, and a line of arithmetic.
The argument is the one that proves the axis of an isosceles triangle is perpendicular to its base: in triangle ABD, with AB=AD, the segment from the apex A to the midpoint M of the base is perpendicular to the base and bisects the apex angle. A rhombus is two such triangles back to back, ABD and CBD, and the diagonal AC is the axis of both.
The conclusion, frozen
The right-angle mark at M: AC ⟂ BD.
Dragging the Figure
Drag D round A: the rhombus opens or closes like a hinge, its diagonals change length in opposite directions, one growing as the other shrinks, and the readout keeps both angles at M at 90°. When the diagonals become equal the rhombus is a square. Drag A or B to move the whole figure or change the side.
D cannot leave its circle, since that would make AD=AB and the figure would no longer be a rhombus. A drag that would flatten the rhombus, with D on the line AB, or carry a vertex out of the frame, is refused.
After dragging D, frozen
D swung down round A: a flatter rhombus, BD short and AC long, still perpendicular.
Consequences
• The angle bisectors. Each diagonal of a rhombus bisects the angles at its two ends; the diagonals are the axes of symmetry of the figure. • Area. The diagonals cut the rhombus into four congruent right triangles, so its area is half the product of the diagonals, 21d1d2. • Side and diagonals. Each right triangle has legs 21d1 and 21d2 and the side of the rhombus as hypotenuse, so by the Pythagorean theorems2=41(d12+d22). • The converse. A parallelogram whose diagonals are perpendicular is a rhombus: the four right triangles round M are congruent by side-angle-side, so the four sides are equal. • Kites. In a kite, with two pairs of equal adjacent sides, one diagonal is still the perpendicular bisector of the other; the rhombus is the kite in which both diagonals have this property, and the square is the rhombus in which the diagonals are also equal.