▶ 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 C; D mirrors B in the axis, and the readout under the canvas keeps the angle at M and the two halves of BD. 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 axis AC splits the kite into two triangles with three equal sides, so by SSS the angles at A are equal: AC bisects the angle at A. Then the two small triangles on either side of the axis agree in two sides and the included angle, so by SAS BM = MD and the two angles at M are equal; they make a straight angle, so each is 90°. Drag A, B or C; D mirrors B.Kite diagonals
Display
Drag A, B or C; the fourth vertex D follows
∠AMB = 90.0°, BM = 2.2, MD = 2.2
Step 1 / 6
Step explanations
1A kite: two pairs of equal adjacent sides
Given ABCD is a kite: AB = AD (one tick each) and CB = CD (two ticks each). The equal sides meet at A and at C, the ends of the axis. To prove: AC ⟂ BD, and AC bisects BD.
Step through the proof, then drag the vertices and watch the diagonals stay at right angles and the crossing stay at the midpoint of the short diagonal. 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?Kite diagonalsDisplayLabelsMeasuresDrag A, B or C; the fourth vertex D follows
∠AMB = 90.0°, BM = 2.2, MD = 2.2
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1A kite: two pairs of equal adjacent sides
Given ABCD is a kite: AB = AD (one tick each) and CB = CD (two ticks each). The equal sides meet at A and at C, the ends of the axis. To prove: AC ⟂ BD, and AC bisects BD.
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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 C move freely; the hollow vertex D is the mirror image of B in the axis AC
4Readout — the line under the figure: the angle at M and the two halves of BD, 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?Kite diagonalsDisplayLabelsMeasuresDrag A, B or C; the fourth vertex D follows
∠AMB = 90.0°, BM = 2.2, MD = 2.2
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1A kite: two pairs of equal adjacent sides
Given ABCD is a kite: AB = AD (one tick each) and CB = CD (two ticks each). The equal sides meet at A and at C, the ends of the axis. To prove: AC ⟂ BD, and AC bisects BD.
2ABC ≅ ADC
SSS congruence Draw the axis AC. The triangles ABC and ADC have AB = AD, CB = CD, and the side AC in common: three pairs of equal sides, so they are congruent by SSS.
3∠BAC = ∠DAC = α
Corresponding parts Congruent triangles have equal corresponding angles. The angle at A in ABC corresponds to the angle at A in ADC, so ∠BAC = ∠DAC: the axis bisects the angle at A. (The same holds at C.)
4ABM ≅ ADM
SAS congruence Draw the other diagonal BD; it meets AC at M. The triangles ABM and ADM have AB = AD, the side AM in common, and the equal angles α between those sides. By SAS they are congruent.
5BM = MD and ∠AMB = ∠AMD = 90°
Linear pair Corresponding parts of ABM and ADM: BM = MD (three ticks), so M is the midpoint of BD; and ∠AMB = ∠AMD = φ. The two angles at M lie along the straight line BD, so φ + φ = 180° and φ = 90°.
6AC ⟂ BD, and AC bisects BD
Conclusion The diagonals of a kite are perpendicular, and the axis AC cuts the other diagonal in half at M. The axis also bisects the angles at A and at C. ∎
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1The axis — AC, joining the two vertices where the equal sides meet; it bisects the angles at A and C
2The crossing M — the midpoint of BD, with the right-angle mark; the two angles φ are the linear pair
3The two halves — BM and MD, three ticks each: corresponding sides of ABM and ADM
4Justification tags — Given, SSS congruence, Corresponding parts, SAS congruence, Linear pair, Conclusion
5Replay — at the last step Play becomes Replay and Next is off
The Statement
A kite is a quadrilateral with two pairs of equal adjacent sides: AB=AD and CB=CD, the equal sides meeting at the two ends A and C of its axis. The theorem says what its diagonals do:
AC⊥BDandBM=MD
where M is the crossing of the diagonals: the diagonals are perpendicular, and the axis cuts the other diagonal in half. The axis also bisects the angles at A and at C. A rhombus is a kite with all four sides equal, and its perpendicular diagonals are the special case.
In the player A, B and C are free; D is placed as the mirror image of B in the line AC, which is exactly what makes AD=AB and CD=CB. The readout under the canvas shows the angle at M and the two halves of BD.
Opening scene, frozen
The kite ABCD: AB = AD (one tick), CB = CD (two ticks).
The Axis and Two Triangles
Draw the axis AC. It splits the kite into the triangles ABC and ADC, which share the side AC and have the kite's equal sides as their other sides:
AB=AD,CB=CD,AC=AC
Three pairs of equal sides, so by the SSS congruence criterion ABC≅ADC. The player shades the two triangles in different colours.
The axis and two triangles, frozen
The axis AC splits the kite into ABC and ADC, congruent by SSS.
The Axis Bisects the Angle
Congruent triangles have equal corresponding angles. In ABC and ADC the angle at A corresponds to the angle at A, so
∠BAC=∠DAC=α
the axis bisects the angle of the kite at A. The same argument at the other end gives ∠BCA=∠DCA. The proof only needs the equal angles at A, which it uses next as the included angles of two smaller triangles.
The axis bisects the angle, frozen
Corresponding angles: ∠BAC = ∠DAC = α.
Two Smaller Triangles
Draw the other diagonal BD and call its crossing with the axis M. The axis now splits the upper part of the kite into the triangles ABM and ADM:
AB=AD,∠BAM=∠DAM=α,AM=AM
Two sides and the included angle agree, so ABM≅ADM by side-angle-side. Nothing new was assumed: the equal sides are the kite's, the equal angles came from the previous step, and AM is shared.
Two smaller triangles, frozen
The diagonal BD and its crossing M; ABM ≅ ADM by SAS.
The Linear Pair
Corresponding parts of ABM and ADM give two things at once. The sides opposite α are equal, BM=MD (three ticks), so M is the midpoint of BD: the axis bisects the other diagonal. And the angles at M are equal, ∠AMB=∠AMD=φ.
Those two angles sit side by side along the straight line BD, a linear pair, so they add up to a straight angle, as on the vertical angles page:
φ+φ=180°⟹φ=90°
The diagonals are perpendicular.
The linear pair, frozen
BM = MD (three ticks) and the equal angles φ at M along BD: φ = 90°.
Why It Holds
The proof uses the kite's definition twice: once through side-side-side, to show that the axis bisects the angle at A, and once through side-angle-side, to show that the two halves of the kite on either side of the axis match. Equal angles along a straight line then have to be right angles. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given sides, SSS, corresponding parts, SAS, and a linear pair.
Another way to say it: A and C are both the same distance from B as from D, so each lies on the perpendicular bisector of BD, and the axis is that perpendicular bisector. The proof above is that fact spelled out with triangles.
The conclusion, frozen
AC ⟂ BD, and M is the midpoint of BD.
Dragging the Figure
Drag B: it swings about the axis and D mirrors it, so the kite changes shape while AB=AD and CB=CD stay true; the readout keeps the angle at M at 90° and the two halves of BD equal. Drag A or C to lengthen or tilt the axis.
The player keeps the kite convex: the foot of B on the axis must fall strictly between A and C. A drag that would push it past an end, making a dart (a concave kite) in which the diagonals still cross at right angles but outside the figure, is refused, as is one that carries a vertex out of the frame.
After dragging B, frozen
B swung down and out, D mirroring it: a wider kite, the right angle at M unchanged.
Consequences
• Area of a kite. The diagonals are perpendicular, so the kite is four right triangles with legs along the diagonals, and its area is half the product of the diagonals: 21AC⋅BD. The same formula holds for a rhombus and a square. • The perpendicular bisector. The axis of a kite is the perpendicular bisector of the other diagonal; every point of it is equidistant from B and D. This is the fact behind the circumcentre, where three such bisectors meet. • Kites on these pages. A circle's centre with the two points of contact of the two tangents from an outside point make a kite with two right angles; the centre, the ends of a chord and the foot of the perpendicular from the centre make the same picture in halves. • Tangential quadrilateral. Every convex kite has an inscribed circle touching all four sides, since the axis bisects the angles at A and C and their bisectors meet the other two angle bisectors at one point. • Not every quadrilateral with perpendicular diagonals is a kite. Perpendicular diagonals alone give the area formula but not equal sides; a kite needs one diagonal to bisect the other as well.