▶ 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 round the near side of the circle, P outside it, or O; the readout under the canvas keeps PT2 beside PA⋅PB. 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?Drop the perpendicular from the centre O to the chord AB; its foot M is the midpoint of AB. Then PA · PB = (PM − AM)(PM + AM) = PM² − AM², and by Pythagoras in the right triangles OMP and OMA this equals OP² − OA² = OP² − R². The tangent gives the same number: OT ⟂ PT, so PT² = OP² − OT² = OP² − R². Drag A round the near side of the circle, or move P or O.Secant and tangent
Display
Drag A round the near side of the circle, P outside, O to move the circle
PT² = 11.81, PA · PB = 11.81
Step 1 / 6
Step explanations
1A tangent and a secant from an outside point
Given P lies outside the circle with centre O. From P, the tangent touches the circle at T, and the secant through P cuts it at A (nearer) and B (farther). To prove: PT² = PA · PB.
Step through the proof, then drag the near end of the secant round the circle, or move the outside point, and watch the square of the tangent stay equal to the product of the two secant segments. 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?Secant and tangentDisplayLabelsMeasuresDrag A round the near side of the circle, P outside, O to move the circle
PT² = 11.81, PA · PB = 11.81
← Back▶ PlayNext →ResetNormal ▾Step 1 / 6
Step explanations
1A tangent and a secant from an outside point
Given P lies outside the circle with centre O. From P, the tangent touches the circle at T, and the secant through P cuts it at A (nearer) and B (farther). To prove: PT² = PA · PB.
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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 slides round the near side of the circle, P and O move freely; the hollow points T and B are computed
4Readout — the line under the figure: PT² beside PA · PB, 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?Secant and tangentDisplayLabelsMeasuresDrag A round the near side of the circle, P outside, O to move the circle
PT² = 11.81, PA · PB = 11.81
← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6
Step explanations
1A tangent and a secant from an outside point
Given P lies outside the circle with centre O. From P, the tangent touches the circle at T, and the secant through P cuts it at A (nearer) and B (farther). To prove: PT² = PA · PB.
2M, the foot of O on AB, is the midpoint of AB
Perpendicular bisects the chord Drop the perpendicular from the centre O to the chord AB; call its foot M. The perpendicular from the centre bisects the chord, so AM = MB (one tick each).
3PA · PB = PM² − AM²
Difference of squares P, A, M, B lie on one line in that order, so PA = PM − AM and PB = PM + MB = PM + AM. Their product is (PM − AM)(PM + AM) = PM² − AM².
4PM² − AM² = OP² − R²
Pythagorean theorem OMP and OMA are right-angled at M. By Pythagoras, PM² = OP² − OM² and AM² = OA² − OM². Subtracting, the OM² cancels: PM² − AM² = OP² − OA² = OP² − R², where R = OA is the radius.
5PT² = OP² − R²
Tangent ⟂ radius + Pythagoras The tangent at T is perpendicular to the radius OT, so OTP is right-angled at T, and by Pythagoras PT² = OP² − OT² = OP² − R²: the same number.
6PT² = PA · PB
Conclusion Both equal OP² − R², which depends only on P and the circle (the power of P). So PT² = PA · PB, and every secant from P gives the same product. ∎
2The two right triangles — OMP and OMA on the chord: PM² − AM² = OP² − R²
3The midpoint M — the foot of the perpendicular from O, with AM = MB (one tick each)
4Justification tags — Given, Perpendicular bisects the chord, Difference of squares, Pythagorean theorem, Tangent ⟂ radius + Pythagoras, Conclusion
5Replay — at the last step Play becomes Replay and Next is off
The Statement
Take a point P outside a circle. From P draw a tangent, touching the circle at T, and a secant, cutting the circle at A (the nearer point) and B (the farther). The theorem says
PT2=PA⋅PB
the square of the tangent segment equals the product of the whole secant and its outside part. It is Proposition 36 of Book III of Euclid's Elements, and it is the outside version of the intersecting chords theorem: in both, the product depends only on P and the circle, a number called the power of P.
The proof is Euclid's and uses no similar triangles, only the perpendicular from the centre, a difference of squares and Pythagoras. In the player O and P are free and A slides round the near side of the circle; B is the second point where the line PA meets the circle, and T the point of contact of the tangent from P. The readout under the canvas shows PT2 and PA⋅PB.
Opening scene, frozen
The circle about O, the outside point P, the tangent PT and the secant through A and B.
The Midpoint of the Chord
The secant cuts off the chord AB. Drop the perpendicular from the centre O to this chord and call its foot M. The perpendicular from the centre to a chord bisects the chord, so
AM=MB
marked in the player with one tick each. The point M is the hinge of the proof: it turns the two secant segments PA and PB into a sum and a difference of the same two lengths.
The midpoint of the chord, frozen
OM perpendicular to the chord: M is the midpoint, AM = MB.
A Difference of Squares
The four points lie on one line in the order P, A, M, B, with A between P and M. So
PA=PM−AM,PB=PM+MB=PM+AM
and the product of the two secant segments is a product of a difference and a sum:
PA⋅PB=(PM−AM)(PM+AM)=PM2−AM2
Only squares of lengths remain, and squares of lengths are what Pythagoras delivers.
A difference of squares, frozen
PA = PM − AM and PB = PM + AM, so PA · PB = PM² − AM².
Two Right Triangles
Both PM and AM are legs of right triangles with the right angle at M and the other leg OM: the triangles OMP and OMA, shaded in the player. By the Pythagorean theorem,
PM2=OP2−OM2,AM2=OA2−OM2
Subtracting, OM2 cancels:
PA⋅PB=PM2−AM2=OP2−OA2=OP2−R2
where R=OA is the radius. The right-hand side knows nothing about the secant: only where P is and how big the circle is. So every secant from P gives the same product.
Two right triangles, frozen
OMP and OMA shaded: PM² − AM² = OP² − OA² = OP² − R².
The Tangent
Now the tangent. The tangent is perpendicular to the radius at the point of contact, so the triangle OTP is right-angled at T, and by Pythagoras once more
PT2=OP2−OT2=OP2−R2
the same number as before. The tangent is the secant's limiting case, in which A and B have come together at T and PA⋅PB has become PT⋅PT; the proof treats it separately only because the chord AB has shrunk to a point and has no midpoint construction.
The tangent, frozen
OT ⟂ PT, so PT² = OP² − OT² = OP² − R².
Why It Holds
Both sides equal OP2−R2, so PT2=PA⋅PB. 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, Pythagoras (twice), and tangent perpendicular to radius with Pythagoras again.
The number OP2−R2 is the power of P with respect to the circle. For P outside it is positive and equals the square of the tangent; for P inside it is negative and its absolute value is the product of the two parts of any chord through P, as on the intersecting chords page; for P on the circle it is zero. One quantity, three theorems.
The conclusion, frozen
Both equal the power of P: PT² = PA · PB.
Dragging the Figure
Drag A round the near side of the circle: the secant turns about P, B and M follow, the two segments PA and PB change, and the readout keeps their product equal to PT2, which does not change at all. Drag P to change the power itself, or O to move the circle; the tangent and its point of contact are recomputed.
The player keeps P outside the circle, keeps A the nearer of the two secant points, and refuses a drag that would make the chord shorter than a few units or carry a point out of the frame.
After dragging P and A, frozen
P moved below and the secant turned: different PA and PB, the same product, the same PT².
Consequences
• Two secants. Two secants from P cutting the circle at A, B and at C, D satisfy PA⋅PB=PC⋅PD, since both products equal OP2−R2. Together with the intersecting chords theorem this is the full power-of-a-point theorem. • Two tangents. Both tangents from P have PT2=OP2−R2, so the two tangents from a point are equal; that page proves it directly, and the hypotenuse-leg criterion gives a third route. • Constructing the tangent length. Given PA and PB, the tangent is their geometric mean, PT=PA⋅PB, which is why the tangent-secant figure appears in the classical construction of a mean proportional. • The converse. If a point T on a circle and a secant PAB satisfy PT2=PA⋅PB, then PT is tangent (Euclid III.37); the proof compares with the true tangent from P and uses this theorem. • Radical axis. The points with equal power with respect to two circles form a line, the radical axis; for two intersecting circles it is the line through their common points.