▶ 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 A, B, C or D round the circle, or drag O to move it; P and the feet follow, and the readout under the canvas keeps PA⋅PB, PC⋅PD and R2−OP2 equal. Learn more about why it holds
Proof?Drop the perpendicular from the centre O to the chord AB; its foot M is the midpoint of AB. Then PA · PB = (AM + MP)(AM − MP) = AM² − MP², and by Pythagoras in the right triangles OMA and OMP this equals OA² − OP² = R² − OP², which depends only on P. The same for the chord CD, so PA · PB = PC · PD. Slide the ends of the chords round the circle.Intersecting chords
Display
Drag A, B, C or D round the circle, O to move the circle
PA·PB = 6.21, PC·PD = 6.21, R² − OP² = 6.21
Step 1 / 6
Step explanations
1Two chords crossing inside the circle
Given AB and CD are chords of the circle with centre O, and they cross at P. To prove: PA · PB = PC · PD.
Step through the proof, then slide the ends of the two chords round the circle and watch the two products in the readout stay equal to each other and to a number that depends only on the crossing point. 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?Intersecting chordsDisplayLabelsMeasuresDrag A, B, C or D round the circle, O to move the circle
PA·PB = 6.21, PC·PD = 6.21, R² − OP² = 6.21
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1Two chords crossing inside the circle
Given AB and CD are chords of the circle with centre O, and they cross at P. To prove: PA · PB = PC · PD.
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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 — A, B, C, D slide round the circle and O moves it; the hollow crossing point P and the feet follow
4Readout — the line under the figure: PA·PB, PC·PD and R² − OP², 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?Intersecting chordsDisplayLabelsMeasuresDrag A, B, C or D round the circle, O to move the circle
PA·PB = 6.21, PC·PD = 6.21, R² − OP² = 6.21
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1Two chords crossing inside the circle
Given AB and CD are chords of the circle with centre O, and they cross at P. To prove: PA · PB = PC · PD.
2The foot M of the perpendicular from O is the midpoint of AB
Perpendicular bisects the chord Drop the perpendicular from O to AB, meeting it at M. The perpendicular from the centre bisects the chord, so AM = MB (one tick each). Join O to A and to P.
3PA · PB = AM² − MP²
Difference of squares P lies on the chord between M and one end, say B. Then PA = AM + MP and PB = MB − MP = AM − MP, so PA · PB = (AM + MP)(AM − MP) = AM² − MP².
4AM² − MP² = OA² − OP² = R² − OP²
Pythagorean theorem OMA and OMP are right triangles at M. So AM² = OA² − OM² and MP² = OP² − OM², and subtracting, AM² − MP² = OA² − OP². OA is the radius R, so PA · PB = R² − OP²: it depends on P and the circle, not on the chord.
5PC · PD = R² − OP² as well
Same argument Repeat with the chord CD: the foot N of the perpendicular from O is its midpoint, PC · PD = CN² − NP², and Pythagoras in ONC and ONP gives CN² − NP² = OC² − OP² = R² − OP².
6PA · PB = PC · PD
Equal to the same Both products equal R² − OP², so they are equal to each other. The number R² − OP² is called the power of the point P with respect to the circle; every chord through P is cut by P into two pieces whose product is this number. ∎
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1The crossing point P — the two products PA·PB and PC·PD both equal R² − OP², its power
2The right triangles on AB — OMA and OMP, with M the midpoint of the chord
3The right triangles on CD — ONC and ONP, with N the midpoint of the other chord
4Justification tags — Given, Perpendicular bisects the chord, Difference of squares, Pythagorean theorem, Same argument, Equal to the same
5Replay — at the last step Play becomes Replay and Next is off
The Statement
Two chords AB and CD of a circle cross at a point P inside it. Each chord is cut by P into two pieces, and the theorem, Proposition 35 of Book III of Euclid's Elements, says that the products of the pieces are equal:
PA⋅PB=PC⋅PD
The usual modern proof uses similar triangles: the inscribed angles on the same arcs make PAC and PDB similar. The proof on this page is Euclid's own, which uses no similarity at all; it shows that each product equals R2−OP2, where R is the radius and O the centre, a quantity that depends on P and the circle but not on the chord. That quantity is the power of the point P.
In the player the four chord ends slide round the circle and P, the crossing point, is computed, as are the feet of the perpendiculars from the centre. The readout under the canvas shows the two products and R2−OP2, in the frame's own units squared, live while you drag.
Opening scene, frozen
Two chords AB and CD crossing at the hollow point P.
marked with one tick each. Join O to A and to P. The point M is the reference from which both A and P will be measured; the whole proof is about the two distances AM and MP.
The perpendicular from the centre, frozen
OM ⟂ AB with M the midpoint of the chord; OA and OP joined.
A Difference of Squares
P lies on the chord, on one side of the midpoint M; say between M and B. Then the piece PA is the half-chord plus MP, and the piece PB is the half-chord minus MP:
PA=AM+MP,PB=MB−MP=AM−MP
Their product is a difference of squares:
PA⋅PB=(AM+MP)(AM−MP)=AM2−MP2
If P is on the other side of M the roles of A and B swap and the result is the same. So the product of the two pieces of any chord through P is the square of its half-length minus the square of the distance from its midpoint to P.
A difference of squares, frozen
PA = AM + MP and PB = AM − MP along the lit chord.
Pythagoras Twice
The triangles OMA and OMP both have a right angle at M, filled in two colours. The Pythagorean theorem in each gives
AM2=OA2−OM2,MP2=OP2−OM2
Subtract, and OM2 cancels:
AM2−MP2=OA2−OP2=R2−OP2
since OA is a radius. So PA⋅PB=R2−OP2. Nothing on the right-hand side refers to the chord: it is the same number for every chord through P.
Pythagoras twice, frozen
The right triangles OMA and OMP: AM² − MP² = OA² − OP².
The Other Chord
Now do the same for CD. The foot N of the perpendicular from O is the midpoint of CD (two ticks), P lies on one side of it, so
PC⋅PD=CN2−NP2
and Pythagoras in the right triangles ONC and ONP gives CN2−NP2=OC2−OP2=R2−OP2. The player fills these two triangles in the two remaining colours.
The other chord, frozen
The same construction on CD: N, ON, and the triangles ONC and ONP.
Why It Holds
Both products have been shown equal to the same number, R2−OP2, so
PA⋅PB=PC⋅PD=R2−OP2
The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given chords, the chord bisector theorem, a difference of squares, the Pythagorean theorem, the same argument on the other chord, and equality to the same thing.
The number R2−OP2 is the power of P with respect to the circle. The proof gives more than the theorem asks: not just that the two products are equal, but what they equal. For P at the centre every chord is a diameter cut in half and the power is R2; as P moves out to the circle the power falls to 0. The similar-triangles proof, which uses the inscribed angle theorem to match the angles of PAC and PDB, proves the equality but not the value.
The conclusion, frozen
Both products equal R² − OP², the power of P.
Dragging the Figure
Slide any chord end round the circle: the chord swings, P moves along the other chord, both products change, and they change together, always matching R2−OP2 in the readout. Move P towards the centre by making the chords cross near O and the products grow towards R2; move it out towards the circle and they shrink towards 0. Drag O to move the whole circle.
The player keeps the chords crossing properly inside the circle, with P away from the centre and from the two midpoints so that the right triangles stay visible; a drag that would separate the chords, or bring P too close to O, M or N, is refused.
After dragging B, frozen
B slid down the circle: AB is steeper and P nearer the centre; the two products are larger and still equal.
Consequences
• Outside the circle. For a point P outside the circle and two lines through it cutting the circle at A, B and C, D, the same argument with P beyond the midpoint gives PA⋅PB=PC⋅PD=OP2−R2: the power is positive outside, zero on the circle, negative inside when signed. • The tangent. When a line through an outside point touches the circle, the two points coincide and the product becomes the square of the tangent length: PT2=OP2−R2, which the two tangents page found with Pythagoras directly. • The converse. If two segments AB and CD cross at P with PA⋅PB=PC⋅PD, the four points lie on a circle; this is a standard test for cyclic quadrilaterals. • A diameter through P. Taking the chord through P and O gives the pieces R−OP and R+OP, whose product is R2−OP2 at once. • The radical axis. The points with equal power with respect to two circles form a line perpendicular to the line of centres; for two crossing circles it is the line through their common points.