The Tan-Chord Theorem

The tan-chord theorem links the angle between a tangent and a chord to an angle inside the circle. It is one of the most used results in geometry riders, and its proof is examinable in Grade 11 and Grade 12.

Theorem. The angle between a tangent and a chord, drawn at the point of contact, equals the angle in the alternate segment. (Reason: tan chord theorem)

Proof. $T A$ is a tangent at $A$ , $A B$ is a chord and $C$ is on the circle in the alternate segment. Draw the diameter $A O D$ and join $D B$ .

$\hat{T A D} = 90°$ (tan $⊥$ radius), so $\hat{T A B} = 90° - \hat{B A D}$
$\hat{A B D} = 90°$ ( $∠$ in semi-circle), so $\hat{A D B} = 90° - \hat{B A D}$ (sum of $∠$ s in $△$ )
Therefore $\hat{T A B} = \hat{A D B}$
$\hat{A D B} = \hat{A C B}$ ( $∠$ s in the same seg)

So $\hat{T A B} = \hat{A C B}$ .

Example. If $\hat{T A B} = 64°$ , then $\hat{A C B} = 64°$ and the angle at the centre on chord $A B$ is $128°$ .

The converse

Converse. If a line drawn through the end of a chord makes an angle with the chord equal to the angle in the alternate segment, the line is a tangent. (Reason: converse tan chord theorem)

Example. In $△ A B C$ , $D$ lies on $A C$ with $\hat{A B D} = \hat{A C B}$ . In the circle through $B$ , $C$ and $D$ , the angle between $A B$ and chord $B D$ equals the angle in the alternate segment, so $A B$ is a tangent to that circle.

Use the converse when a question asks you to prove that a line is a tangent.

Key terms

Alternate segment: The segment on the other side of the chord from the angle between the tangent and the chord.
Point of contact: Where the tangent touches the circle.
Converse: The theorem used backwards, to prove a line is a tangent.

Common mistakes

Using the angle in the wrong segment.
Applying the theorem to a line that has not been given or proved to be a tangent.
Leaving out the construction in the proof.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

$T A$ is a tangent at $A$ to a circle with centre $O$ . $A B$ is a chord and $C$ is a point on the circle in the alternate segment. $\hat{T A B} = 64°$ .

Write down $\hat{A C B}$ and calculate $\hat{A O B}$ , with reasons.

Worked solution

Hidden - try the question first

Solution

$\hat{A C B} = 64°$ (tan chord theorem)
$\hat{A O B} = 2 \times 64°$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$= 128°$

Question 2

5 marks total

routine

In the diagram, $T A S$ is a tangent to the circle with centre $O$ at $A$ , $A B$ is a chord and $C$ lies on the circle in the alternate segment. Prove the theorem that the angle between a tangent and a chord is equal to the angle in the alternate segment, that is, $\hat{T A B} = \hat{A C B}$ .

Worked solution

Hidden - try the question first

Solution

Construction: diameter $A O D$ ; join $D B$
$\hat{T A D} = 90°$ (tan $⊥$ radius)
$\hat{A B D} = 90°$ ( $∠$ in semi-circle)
$\hat{T A B} = 90° - \hat{B A D}$ , and $\hat{A D B} = 90° - \hat{B A D}$ (sum of $∠$ s in $△ A B D$ ), so $\hat{T A B} = \hat{A D B}$
$\hat{A D B} = \hat{A C B}$ ( $∠$ s in the same seg), so $\hat{T A B} = \hat{A C B}$

Question 3

6 marks total

complex

$T A$ is a tangent at $A$ . $B$ and $C$ lie on the circle, with $C$ in the alternate segment for chord $A B$ . $\hat{T A B} = 50°$ and $\hat{A B C} = 70°$ .

Calculate, with reasons:

(a) $\hat{A C B}$ (2)

(b) $\hat{B A C}$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $A O D$ (tan chord theorem)
$= 50°$
(b) $\hat{B A C} = 180° - 50° - 70°$ (sum of $∠$ s in $△$ )
$= 60°$
(c) $A C$ , which is opposite the largest angle
$\hat{A B C} = 70°$ is the largest angle in the triangle

Question 4

8 marks total

problem-solving

In the diagram, $D$ lies on side $A C$ of $△ A B C$ so that $\hat{A B D} = \hat{A C B}$ . The circle through $B$ , $C$ and $D$ is drawn.

(a) Prove that $A B$ is a tangent to the circle through $B$ , $C$ and $D$ . (3)

(b) If $\hat{A B D} = 35°$ and $\hat{B D C} = 100°$ , calculate $\hat{A}$ and $\hat{D B C}$ . (5)

Worked solution

Hidden - try the question first

Solution

(a) In the circle through $A O D$ , $C$ and $D$ , $B D$ is a chord
$\hat{A O B} = 2 \times 64°$ , $D$ and $C$ are collinear, so $\hat{A C B} = \hat{D C B}$ , which is in the alternate segment for chord $B D$
$∠$ , so $\hat{A D B} = 90° - \hat{B A D}$ is a tangent (converse tan chord theorem)
(b) $\hat{B D C}$ is an exterior angle of $△ A B D$ : $\hat{B D C} = \hat{A} + \hat{A B D}$
$\hat{A} = 100° - 35° = 65°$
$\hat{A C B} = 35°$
In $T A$ : $\hat{D B C} = 180° - 100° - 35°$
$= 45°$

FAQs

Is the tan-chord theorem proof examinable?

Yes, in Grade 11 and Grade 12. The converse is used as a result, not proved.

How do I find the alternate segment?

Look at the chord at the point of contact. The alternate segment is on the opposite side of that chord from the tangent angle.

How do I prove a line is a tangent?

Show it is perpendicular to a radius at the circle, or use the converse of the tan-chord theorem.

The tan-chord theorem