▶ 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 or C round the near side of the circle, P outside it, or O; the readout under the canvas keeps PA⋅PB beside PC⋅PD. 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?For each secant, drop the perpendicular from the centre O to its chord; the foot is the midpoint, and the product of the two secant segments is PM² − AM², a difference of squares. Pythagoras in the right triangles OMP and OMA turns this into OP² − R², which does not depend on the secant. So both products are the same number, the power of P. Drag A or C round the near side of the circle, or move P or O.Two secants
Display
Drag A or C round the near side, P outside, O to move the circle
PA · PB = 9.89, PC · PD = 9.89
Step 1 / 6
Step explanations
1Two secants from an outside point
Given P lies outside the circle with centre O. Two secants from P cut the circle at A, B and at C, D, with A and C the nearer points. To prove: PA · PB = PC · PD.
Step through the proof, then drag the near end of either secant round the circle, or move the outside point, and watch the two products stay equal. 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?Two secantsDisplayLabelsMeasuresDrag A or C round the near side, P outside, O to move the circle
PA · PB = 9.89, PC · PD = 9.89
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1Two secants from an outside point
Given P lies outside the circle with centre O. Two secants from P cut the circle at A, B and at C, D, with A and C the nearer points. 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, Measures writes each marked angle's size beside its arc
3Draggable points — A and C slide round the near side of the circle, P and O move freely; the hollow points B and D are computed
4Readout — the line under the figure: PA · PB beside PC · PD, 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?Two secantsDisplayLabelsMeasuresDrag A or C round the near side, P outside, O to move the circle
PA · PB = 9.89, PC · PD = 9.89
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1Two secants from an outside point
Given P lies outside the circle with centre O. Two secants from P cut the circle at A, B and at C, D, with A and C the nearer points. To prove: PA · PB = PC · PD.
2M and N, the feet of O, are the midpoints of AB and CD
Perpendicular bisects the chord Drop the perpendiculars from the centre to the two chords; call the feet M and N. The perpendicular from the centre bisects the chord, so AM = MB (one tick each) and CN = ND (two ticks each).
3PA · PB = PM² − AM² and PC · PD = PN² − CN²
Difference of squares On the first secant, PA = PM − AM and PB = PM + AM, so PA · PB = (PM − AM)(PM + AM) = PM² − AM². On the second, in the same way, PC · PD = PN² − CN².
4PM² − AM² = OP² − R²
Pythagorean theorem OMP and OMA are right-angled at M. By Pythagoras, PM² = OP² − OM² and AM² = OA² − OM². Subtracting, OM² cancels: PA · PB = PM² − AM² = OP² − OA² = OP² − R², with R = OA the radius.
5PN² − CN² = OP² − R² as well
Pythagorean theorem The same two right triangles on the other secant: ONP and ONC are right-angled at N, so PN² = OP² − ON² and CN² = OC² − ON², and PC · PD = PN² − CN² = OP² − OC² = OP² − R². Nothing in this number says which secant was used.
6PA · PB = PC · PD
Conclusion Both products equal OP² − R², the power of P with respect to the circle. So PA · PB = PC · PD, and every secant from P gives the same product. ∎
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1The first secant — OMP and OMA on the chord AB: PM² − AM² = OP² − R²
2The second secant — ONP and ONC on the chord CD: PN² − CN² = OP² − R²
3The midpoints — M and N, feet of the perpendiculars from O, with one and two ticks for the equal halves
4Justification tags — Given, Perpendicular bisects the chord, Difference of squares, Pythagorean theorem (twice), Conclusion
5Replay — at the last step Play becomes Replay and Next is off
The Statement
Take a point P outside a circle and draw two secants from it, one cutting the circle at A and B, the other at C and D, with A and C the nearer points. Then
PA⋅PB=PC⋅PD
the product of the whole secant and its outside part is the same for every secant from P. This completes the power-of-a-point family on these pages: the intersecting chords theorem is the same statement for a point inside the circle, and the secant-tangent theorem is the case where one secant has shrunk to a tangent. In Euclid it is the outside half of Proposition 36 of Book III.
The proof is the secant-tangent argument done twice. In the player O and P are free, A and C slide round the near side of the circle, and B, D are where the lines PA, PC meet the circle again. The readout under the canvas shows the two products.
Opening scene, frozen
The circle about O, the outside point P, and the two secants through A, B and C, D.
The Two Midpoints
Each secant cuts off a chord, AB and CD. Drop the perpendicular from the centre O to each chord and call the feet M and N. The perpendicular from the centre bisects the chord, so
AM=MB,CN=ND
marked with one tick each on the first chord and two ticks each on the second. The feet M and N are where the two secants will be compared: each of them turns its secant's two segments into a sum and a difference of the same two lengths.
The two midpoints, frozen
OM and ON perpendicular to the chords: AM = MB, CN = ND.
Differences of Squares
On the first secant the points lie in the order P, A, M, B, so PA=PM−AM and PB=PM+MB=PM+AM, and
PA⋅PB=(PM−AM)(PM+AM)=PM2−AM2
On the second secant, with P, C, N, D in order, the same algebra gives
PC⋅PD=(PN−CN)(PN+CN)=PN2−CN2
Two products have become two differences of squares, and squares of lengths are what Pythagoras provides.
Differences of squares, frozen
PA · PB = PM² − AM² and PC · PD = PN² − CN².
The First Secant
PM and AM are legs of the right triangles OMP and OMA, both right-angled at M with the common leg OM; the player shades them. By the Pythagorean theorem,
PM2=OP2−OM2,AM2=OA2−OM2
Subtracting, OM2 cancels and only the distance to the centre and the radius remain:
PA⋅PB=PM2−AM2=OP2−OA2=OP2−R2
The first secant, frozen
OMP and OMA shaded: PM² − AM² = OP² − R².
The Second Secant
Nothing in the previous step used anything special about the first secant. Repeat it on the second: ONP and ONC are right-angled at N with the common leg ON, so PN2=OP2−ON2 and CN2=OC2−ON2, and
PC⋅PD=PN2−CN2=OP2−OC2=OP2−R2
The foot moved from M to N and the radius from OA to OC, but the result is the same expression, because every radius is R and OP is the same segment.
The second secant, frozen
ONP and ONC shaded: PN² − CN² = OP² − R² as well.
Why It Holds
Both products equal OP2−R2, so PA⋅PB=PC⋅PD. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, the perpendicular bisecting the chord, a difference of squares, and Pythagoras twice on each secant.
The number OP2−R2 is the power of P with respect to the circle. It is positive outside the circle, where it is the square of the tangent and the product of any secant's segments; negative inside, where its absolute value is the product of the two parts of any chord through P; and zero on the circle. Three theorems, one quantity, one proof pattern: midpoint, difference of squares, Pythagoras.
The conclusion, frozen
Both products are the power of P: PA · PB = PC · PD.
Dragging the Figure
Drag A or C round the near side of the circle: its secant turns about P, the far end and the midpoint follow, the two segments change, and the readout keeps the product equal to the other secant's product. Drag P to change the power itself, or O to move the circle.
The player keeps P outside the circle, keeps A and C the nearer ends of their secants, keeps the two secants apart, and refuses a drag that would make a chord shorter than a few units or carry a point out of the frame.
After dragging P, A and C, frozen
P moved below and both secants turned: four different segments, the two products still equal.
Consequences
• Tangents as a limit. Let C and D come together: the second secant becomes a tangent PT and the product becomes PT2, which is the secant-tangent theorem; let both become tangents and the two tangents from P are equal. • Four concyclic points. The converse holds: if two lines through P meet at A, B and C, D with PA⋅PB=PC⋅PD and P outside both segments, then A, B, C, D lie on one circle. This is the usual test for a cyclic quadrilateral by lengths rather than angles. • Similar triangles. Joining A to D and C to B makes the triangles PAD and PCB similar, with the common angle at P and equal inscribed angles at B and D; the equal products are the proportion PA:PC=PD:PB read crosswise. These pages avoid similarity, which is why the proof above goes through Pythagoras instead. • Radical axis. For two circles, the points with equal power with respect to both lie on a line; a point on it sends equal tangents to both circles. • Measuring an unreachable circle. The power of P can be found from any one secant, and then every other secant from P is pinned down: given the outside part, the whole follows.