Angles in the same segment and cyclic quadrilaterals – Circle and Related Theorems (Senior 3 Mathematics) | Ganzaa

Circle and Related Theorems

Lessons

Senior 3 · Mathematics

Circle and Related Theorems

5 lessons

Lesson 4 of 5

Angles in the same segment and cyclic quadrilaterals

25 min

Steering by an angle

At night, the skipper of a boat on a lake sees two lights on the shore, $A$ and $B$ , and measures the angle between them with a sighting compass. That single angle does not fix the boat's position, but it puts the boat somewhere on one particular arc through $A$ and $B$ : every point of that arc sees $A B$ at the same angle. Sailors have long used this "danger angle" to keep clear of rocks. The fact behind it is the first theorem of this lesson.

Definition

A chord $A B$ cuts a circle into two segments. Angles $\angle ACB$ , $\angle ADB$ , … with $C$ , $D$ , … all in the same segment are called angles in the same segment.

A cyclic quadrilateral is a quadrilateral whose four vertices lie on one circle. Points that lie on one circle are concyclic.

Example: in the figure below, $C$ , $D$ and $E$ are all in the major segment cut off by $A B$ , so $A C D E$ , for example, is a cyclic quadrilateral.

Important

Theorem 3: angles in the same segment are equal. In the figure, $\angle ACB = \angle ADB = \angle AEB$ .

Proof. Each of them equals $\frac{1}{2}\angle AOB$ (angle at the centre is twice the angle at the circumference), so they are equal to each other. Here $\angle AOB = 100^\circ$ and each angle is $50^\circ$ .

Worked example 1: same segment

$A$ , $B$ , $C$ and $D$ lie on a circle with centre $O$ , and $C$ and $D$ are on the same side of $A B$ . $\angle ADB = 38^\circ$ . Find $\angle ACB$ and $\angle AOB$ .

Statement	Reason
$\angle ACB = 38^\circ$	(angles in the same segment)
$\angle AOB = 2 \times 38^\circ = 76^\circ$	(angle at the centre is twice the angle at the circumference)

The cyclic quadrilateral

Important

Theorem 4: opposite angles of a cyclic quadrilateral add up to $180^\circ$ . In a cyclic quadrilateral $A B C D$ : $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$ .

Proof (figure above). Let $\angle BAD = x$ and $\angle BCD = y$ .

Statement	Reason
The angle $B O D$ facing $C$ is $2 x$	(angle at the centre is twice the angle at the circumference)
The angle $B O D$ facing $A$ is $2 y$	(the same theorem)
$2x + 2y = 360^\circ$	(angles at a point)
$x + y = 180^\circ$	(dividing by 2)

Two consequences.

An exterior angle of a cyclic quadrilateral equals the interior opposite angle. (If $D C$ is extended to $E$ , then $\angle BCE = 180^\circ - \angle BCD = \angle BAD$ .)
Converse: if a pair of opposite angles of a quadrilateral add up to $180^\circ$ , the quadrilateral is cyclic.

Worked example 2: all four angles

$A B C D$ is a cyclic quadrilateral with $\angle A = 100^\circ$ and $\angle B = 110^\circ$ . Find $\angle C$ and $\angle D$ .

Statement	Reason
$x = \angle C = 180^\circ - 100^\circ = 80^\circ$	(opposite angles of a cyclic quadrilateral)
$y = \angle D = 180^\circ - 110^\circ = 70^\circ$	(opposite angles of a cyclic quadrilateral)
Check: $100^\circ + 110^\circ + 80^\circ + 70^\circ = 360^\circ$	(angle sum of a quadrilateral)

Worked example 3: an equation from the theorem

In a cyclic quadrilateral $A B C D$ , $\angle ABC = 3x$ and $\angle ADC = 2x$ . Find both angles.

Statement	Reason
$3x + 2x = 180^\circ$	(opposite angles of a cyclic quadrilateral)
$x = 36^\circ$	(solving)
$\angle ABC = 108^\circ$ , $\angle ADC = 72^\circ$	(substituting)

Worked example 4: from the centre to the quadrilateral

$A$ , $B$ , $C$ and $D$ lie on a circle with centre $O$ , and $\angle BOD = 140^\circ$ on the side of $C$ . Find $\angle BAD$ and $\angle BCD$ .

Statement	Reason
$\angle BAD = \frac{140^\circ}{2} = 70^\circ$	(angle at the centre is twice the angle at the circumference: both stand on the arc $B C D$ )
$\angle BCD = 180^\circ - 70^\circ = 110^\circ$	(opposite angles of a cyclic quadrilateral)

Two chords that cross

Important

Theorem 5: intersecting chords. If two chords $A B$ and $C D$ of a circle cross at a point $P$ inside the circle, then

PA \times PB = PC \times PD .

Proof. Draw $A C$ and $D B$ .

Statement	Reason
$\angle PAC = \angle PDB$	(angles in the same segment: both stand on the arc $C B$ )
$\angle APC = \angle DPB$	(vertically opposite angles)
Triangles $P A C$ and $P D B$ are similar	(two pairs of equal angles, so the third pair is equal too)
$\frac{PA}{PD} = \frac{PC}{PB}$ , so $PA \times PB = PC \times PD$	(sides of similar triangles are in the same ratio)

Worked example 5: crossing chords

In the figure above, $P A = 4$ cm, $P B = 6$ cm and $P C = 3$ cm. Find $P D = x$ and the length of the chord $C D$ .

Solution $3x = 4 \times 6 = 24$ (intersecting chords), so $x = 8$ cm, and $C D = 3 + 8 = 11$ cm.

Warning

The same segment means the same side of the chord. Two angles on $A B$ from opposite sides are not equal: they add up to $180^\circ$ , because together with $A$ and $B$ they make a cyclic quadrilateral.

A quadrilateral is cyclic only if all four vertices are on the circle. If one vertex is the centre $O$ , Theorem 4 does not apply.

Tip

To spot equal angles in the same segment, find the chord both angles "look at", then check that the two vertices are on the same side of it. Highlighting the chord in colour makes the pairs jump out.

Corbettmaths: Angles in a cyclic quadrilateral proof. Watch how the two angles at the centre fill a full turn, the step used in the proof of Theorem 4.

✦

Practice

Give a reason for each step. The last two are harder.

$C$ and $D$ are on the same side of the chord $A B$ , and $\angle ACB = 42^\circ$ . Find $\angle ADB$ .
$P Q R S$ is a cyclic quadrilateral with $\angle P = 73^\circ$ and $\angle Q = 118^\circ$ . Find $\angle R$ and $\angle S$ .
$A B C D$ is a cyclic quadrilateral, and $A B$ is extended beyond $B$ to $E$ . If $\angle CBE = 65^\circ$ , find $\angle ADC$ .
In a cyclic quadrilateral, $\angle ABC = (2x + 10)^\circ$ and $\angle ADC = (3x - 5)^\circ$ . Find $x$ and both angles.
A quadrilateral has angles $88^\circ$ , $95^\circ$ , $92^\circ$ and $85^\circ$ , in order round the shape. Is it cyclic? Explain.
Chords $A B$ and $C D$ cross at $P$ . $P A = 5$ cm, $P B = 8$ cm and $P C = 4$ cm. Find $P D$ .
Chords $A B$ and $C D$ cross at $P$ , with $P A = 3$ cm, $P B = 12$ cm and $C D = 13$ cm. Find $P C$ and $P D$ .
A chord 12 cm long is cut at right angles by a diameter, at a point 2 cm from one end of the diameter. Find the radius of the circle.

Show answersHide answers

$42^\circ$ (angles in the same segment).
$\angle R = 107^\circ$ , $\angle S = 62^\circ$ (opposite angles of a cyclic quadrilateral).
$\angle ADC = \angle CBE = 65^\circ$ (exterior angle of a cyclic quadrilateral equals the interior opposite angle).
$5 x + 5 = 180$ (opposite angles), so $x = 35$ ; the angles are $80^\circ$ and $100^\circ$ .
Yes: $88^\circ + 92^\circ = 180^\circ$ and $95^\circ + 85^\circ = 180^\circ$ , so by the converse of Theorem 4 it is cyclic.
$4 \times PD = 5 \times 8$ (intersecting chords), so $P D = 10$ cm.
Let $P C = x$ , so $P D = 13 - x$ : $x (13 - x) = 36$ , $x^{2} - 13 x + 36 = 0$ , $(x - 4) (x - 9) = 0$ . So $P C$ and $P D$ are 4 cm and 9 cm (in either order).
The diameter bisects the chord (chord theorem), so the chord's two parts are 6 cm each. If the diameter's parts are 2 cm and $y$ : $2y = 6 \times 6$ (intersecting chords), $y = 18$ . The diameter is 20 cm, so the radius is 10 cm.

Sources and credits

Written by Ganzaa.
Video: Angles in a Cyclic Quadrilateral Proof, Corbettmaths, embedded from YouTube.

The diagrams, proofs, worked examples and practice items were written by Ganzaa. Every angle and length in the diagrams is drawn to its stated size.

1Steering by an angle

2The cyclic quadrilateral

3Two chords that cross

Steering by an angle

The cyclic quadrilateral

Two chords that cross