Alternate segment theorem

A circle touched by a tangent at point $T$, with a chord $TC$ drawn from $T$, is divided by that chord into two segments — two bowl-shaped regions of the circle's interior, one on each side of $TC$. The tangent at $T$ also makes two angles with the chord, one on each side. The theorem pairs each tangent-chord angle with the inscribed angle from the segment lying on the far side of the chord — the "alternate" one, not the near one.

Statement. If $PT$ is tangent to a circle at $T$, $TC$ is a chord, and $A$ is any point on the arc lying in the segment on the opposite side of $TC$ from the angle $\angle PTC$, then

where: $\angle PTC$ is the tangent-chord angle measured between the tangent ray $PT$ and the chord $TC$, and $\angle TAC$ is the inscribed angle that the same chord $TC$ subtends at any point $A$ on the arc of the alternate segment. Two things make this statement easy to misuse if read carelessly. First, there are two tangent-chord angles at $T$, one on each side of $TC$, and they are supplementary (they add to $180^\circ$ since $P$, $T$, and the tangent's continuation form a straight line). Each of these two angles pairs with the inscribed angle of a different segment — pick the wrong side and you will use the supplement instead of the actual answer. Second, "alternate" specifically means the segment cut off on the far side of the chord from where the angle is measured, exactly mirroring the language of alternate angles between parallel lines: the two regions are on opposite sides of a common line (here, the chord).

Prove that the tangent-chord angle $\angle PTC$ at the point of contact $T$ equals the inscribed angle $\angle TAC$ subtended by the same chord $TC$ at any point $A$ on the arc of the alternate segment, using only the tangent-radius perpendicularity fact, the isosceles triangle formed by two radii, and the angle-at-centre theorem.

Step 1. Let $O$ be the centre of the circle and draw the radius $OT$ to the point of contact. Since $PT$ is tangent at $T$, the tangent-radius property gives $\angle OTP = 90^\circ$.

Step 2. Suppose the tangent-chord angle is $\angle PTC = x$. Since $\angle OTP = 90^\circ$ splits into $\angle OTP = \angle OTC + \angle CTP$, the angle between the radius $OT$ and the chord is $\angle OTC = 90^\circ - x$.

Step 3. Triangle $OTC$ is isosceles because $OT = OC$ (both radii), so the base angles are equal: $\angle OCT = \angle OTC = 90^\circ - x$.

Step 4. Using the angle sum of triangle $OTC$, the angle at the centre facing chord $TC$ is

Step 5. By the angle-at-centre theorem, the inscribed angle $\angle TAC$ at any point $A$ on the major arc (the alternate segment, on the far side of $TC$ from the tangent-chord angle) equals half the central angle standing on the same chord:

Step 6. Since $\angle PTC = x$ from Step 2 and $\angle TAC = x$ from Step 5, both angles equal the same quantity $x$, so $\angle PTC = \angle TAC$ — the alternate segment theorem. The proof goes through symmetrically for the tangent-chord angle on the other side of the chord, pairing it with the inscribed angle in the other (minor-arc) segment.

$PT$ is tangent to a circle at $T$, and $TA$ is a chord with the tangent-chord angle $\angle PTA = 65^\circ$. $B$ lies on the arc in the segment alternate to $\angle PTA$, forming triangle $TAB$ with $\angle ATB = 48^\circ$. Find $\angle TBA$ and $\angle TAB$.

Step 1. Apply the alternate segment theorem directly: the tangent-chord angle equals the inscribed angle in the alternate segment, and $B$ was given as lying in that segment for chord $TA$. So

Step 2. Use the angle sum of triangle $TAB$, since $T$, $A$, $B$ are its three vertices:

Step 3. Substitute the known values $\angle ATB = 48^\circ$ and $\angle TBA = 65^\circ$ from Step 1:

So $\angle TBA = \boxed{65^\circ}$ and $\angle TAB = \boxed{67^\circ}$.

Competitive stretch

Suppose instead only $\angle TAB = 67^\circ$ and $\angle TBA = 65^\circ$ were given, with $B$ still in the alternate segment for chord $TA$. Working backward, the alternate segment theorem still gives the tangent-chord angle $\angle PTA = \angle TBA = 65^\circ$ directly — the theorem runs both ways, so an inscribed angle can just as easily hand you an unknown tangent-chord angle as the reverse.

• Pairing the tangent-chord angle with the inscribed angle on the same side of the chord instead of the alternate (opposite) side — this gives the supplement of the correct angle, not the correct angle itself.
• Confusing the tangent-chord angle (between tangent and chord) with the tangent-radius angle (always exactly $90^\circ$) — these are two different, unrelated facts that both involve the same tangent line.
• Forgetting that two tangent-chord angles exist at the point of contact, one on each side of the chord, each linked to a different alternate segment — always check which side the given angle is measured on before matching it to a point.
• Applying the theorem with no tangent present, using it on a configuration that only has two ordinary chords — the alternate segment theorem strictly requires a tangent meeting a chord at the point of contact.

Ask the tutor: Ask the AI tutor to sketch your specific tangent-chord diagram and point out exactly which segment is "alternate" if you're stuck locating it.