GeometryDifficulty 30-8031 tagged problems

Tangent-Chord Theorem

Tangent-Chord Theorem is the habit of converting tangent angles into inscribed angles to unlock circle configurations. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

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The Key Result

$$\angle(\text{tangent}, \text{chord}) = \tfrac{1}{2}\,(\text{intercepted arc}) = \angle(\text{inscribed in alternate segment})$$

The angle between a tangent and a chord equals the inscribed angle on the far side of that chord (the alternate segment).

Why It Works

1
Let a tangent touch a circle of center $O$ at $A$ and let $AB$ be a chord. The tangent-chord angle $\angle(t, AB)$ intercepts arc $AB$ on the side it opens toward.
2
The radius $OA$ is perpendicular to the tangent at $A$, so the tangent-chord angle is $90^\circ - \angle OAB$. Triangle $OAB$ is isosceles ($OA = OB = r$), giving $\angle OAB = \tfrac{1}{2}(180^\circ - \angle AOB) = 90^\circ - \tfrac{1}{2}\angle AOB$. Hence the tangent-chord angle $= 90^\circ - \angle OAB = \tfrac{1}{2}\angle AOB = \tfrac{1}{2}\,\text{arc } AB$.
3
Any inscribed angle subtending the same arc $AB$ from the alternate segment is also half that arc.
4
Therefore the tangent-chord angle equals the inscribed angle in the alternate segment — letting you replace an awkward tangent angle with a normal triangle angle.

Worked Examples

Example 1

A line is tangent to a circle at point $A$, and $AB$ is a chord. A point $C$ lies on the major arc $AB$. The tangent-chord angle between the tangent and $AB$ is $50^\circ$. Find the inscribed angle $\angle ACB$.

1
$C$ lies in the alternate segment relative to the chord $AB$ and the tangent.
2
By the tangent-chord (alternate segment) theorem, $\angle ACB$ equals the tangent-chord angle.
3
Therefore $\angle ACB = 50^\circ$.

Answer:

$\angle ACB = 50^\circ$

Warm-up

A tangent touches a circle at $A$, and $AB$ is a chord with the minor arc $AB$ measuring $84^\circ$. Find the acute tangent-chord angle between the tangent and $AB$.

1
The tangent-chord angle equals half the intercepted arc.
2
The acute angle intercepts the minor arc $AB = 84^\circ$, so the angle is $\tfrac12\cdot 84^\circ = 42^\circ$.

Answer:

$42^\circ$

Contest level

Circle $\omega$ is tangent to line $\ell$ at $A$. Chord $AB$ makes a tangent-chord angle of $35^\circ$ with $\ell$, and chord $AC$ makes a tangent-chord angle of $80^\circ$ with $\ell$ on the same side. Find the inscribed angle $\angle BAC$ and the arc $BC$ they cut.

1
Each tangent-chord angle equals the inscribed angle in the alternate segment, so the angle between the two chords at $A$ is $\angle BAC = 80^\circ - 35^\circ = 45^\circ$ (chords on the same side of the tangent).
2
The inscribed angle $\angle BAC$ subtends arc $BC$, so $\text{arc } BC = 2\angle BAC = 90^\circ$.

Answer:

$\angle BAC = 45^\circ$ and arc $BC = 90^\circ$.

Olympiad / Challenge

Two circles are tangent internally at $T$. A chord $AB$ of the larger circle is tangent to the smaller circle at $P$. Prove that $TP$ bisects $\angle ATB$.

1
Because the circles are internally tangent at $T$, the homothety centered at $T$ taking the small circle to the large one fixes the common tangent at $T$ and maps tangent lines to parallel tangent lines.
2
This homothety sends the tangent point $P$ of the small circle to the point $M$ of the large circle where the tangent is parallel to line $AB$ — namely the midpoint of arc $AB$ not containing $T$.
3
Therefore $T$, $P$, $M$ are collinear, where $M$ is the midpoint of arc $AB$ not containing $T$.
4
Since $M$ is the arc midpoint, $TM$ bisects $\angle ATB$ (inscribed angles on equal arcs $AM$, $MB$); as $P$ lies on $TM$, $TP$ bisects $\angle ATB$.

Answer:

$TP$ bisects $\angle ATB$ (Archimedes' lemma / the shooting lemma).

Going Deeper

Generalization: the tangent-chord angle is the limiting case of the inscribed angle as one endpoint of the chord slides into the tangent point. So 'tangent-chord = inscribed in alternate segment' is just the inscribed-angle theorem extended to a degenerate chord — no new rule, one continuous family.
Where it appears: this is the key to mixtilinear-incircle, tangent-circle, and 'shooting lemma' configurations (Archimedes' lemma above). Whenever a tangent line and a chord share a point, replace the tangent-chord angle with an ordinary triangle angle and the problem becomes a normal angle chase.
Pitfall: a tangent-chord angle has TWO values that sum to $180^\circ$ (one on each side of the chord), matching the two arcs. Use the one whose intercepted arc is on the side the angle opens toward; grabbing the wrong side gives the supplement and a wrong answer.

Spot the Signal

Look for problems where the key step is converting tangent angles into inscribed angles to unlock circle configurations.
You can describe the hard part as converting tangent angles into inscribed angles to unlock circle configurations, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the diagram relation that calls for tangent-chord theorem, then rewrite the givens around it.
Name the relation that makes Tangent-Chord Theorem legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Tangent-Chord Theorem just because the surface looks familiar; verify the required condition first.
Applying Tangent-Chord Theorem because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let tangent-chord theorem reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is converting tangent angles into inscribed angles to unlock circle configurations.