The inscribed angle

Geometry, grade 10 (5 units) Lesson 20 of 26Learning

The inscribed angle

An inscribed angle is an angle whose vertex is on the circle and whose arms cut the circle.

$\displaystyle \angle ACB$ is an inscribed angle: its vertex $\displaystyle C$ is on the circle, and its arms are the chords $\displaystyle CA$ and $\displaystyle CB$ .
It stands on the arc $\displaystyle AB$ inside it – the arc without $\displaystyle C$ .
The central angle $\displaystyle \angle AOB$ stands on the same arc.

The inscribed angle and the central angle

Here the centre

\displaystyle O

is inside the angle

\displaystyle ACB

. When it is on an arm of the angle, or outside it, the proof uses the same diameter

\displaystyle CD

, and in the last step one angle is taken from the other instead of adding them.

Theorem: An inscribed angle is half the central angle standing on the same arc.

Given: $\displaystyle O$ is the centre of the circle; $\displaystyle A$ , $\displaystyle B$ , $\displaystyle C$ are on the circle.
To prove: $\displaystyle {\angle AOB}={2\angle ACB}$ .

The proof below draws the diameter $\displaystyle CD$ . On each side of it, an isosceles triangle with two radii as sides and its exterior angle at $\displaystyle O$ show that the central angle is twice the inscribed one:

	Claim	Reason
1	$\displaystyle {OA}={OC}$	Radii of the circle
2	$\displaystyle {\angle OAC}={\angle OCA}$	(1); the base angles of an isosceles triangle are equal
3	$\displaystyle {\angle AOD}={\angle OAC+\angle OCA}$	An exterior angle of the triangle $\displaystyle AOC$ equals the sum of the two interior angles not next to it
4	$\displaystyle {\angle AOD}={2\angle ACO}$	(2), (3)
5	$\displaystyle {\angle BOD}={2\angle BCO}$	In the same way, in the isosceles triangle $\displaystyle BOC$
6	$\displaystyle {\angle AOB}={2\angle ACB}$	(4), (5); adding: $\displaystyle {\angle AOB}={\angle AOD+\angle BOD}$ and $\displaystyle {\angle ACB}={\angle ACO+\angle BCO}$

An arc is measured by its central angle, so: an inscribed angle is half the arc it stands on.

\displaystyle \angle ACB

\displaystyle =

\displaystyle \frac{1}{2}\angle AOB

Inscribed angles on the same arc

Every inscribed angle is half its arc, so:

inscribed angles standing on the same arc are equal – each is half the same central angle;
in a circle, equal inscribed angles stand on equal arcs and on equal chords;
in a circle, equal arcs – and equal chords – have equal inscribed angles standing on them, from the same side.

Exercise 1

$\displaystyle O$ is the centre of the circle, and $\displaystyle {\angle AOB}={130^{\circ}}$ .
$\displaystyle C$ and $\displaystyle D$ are on the circle, on the same side of $\displaystyle AB$ .

Find

\displaystyle \angle ACB

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle {\angle CAD}={20^{\circ}}

. Find the central angle

\displaystyle \angle COD

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

An inscribed angle is $\displaystyle (x+20)^{\circ}$ , and the central angle on the same arc is $\displaystyle (3x)^{\circ}$ .
Find $\displaystyle x$ .

The inscribed angle on a diameter

Theorem: An inscribed angle standing on a diameter is a right angle.

The central angle on a diameter is a straight angle, $\displaystyle 180^{\circ}$ , and the inscribed angle is half of it.

\displaystyle \angle ACB

\displaystyle =

\displaystyle 90^{\circ}

The converse theorem: an inscribed angle of $\displaystyle 90^{\circ}$ stands on a diameter.

In the right-angled triangle $\displaystyle ACB$ , the median $\displaystyle CM$ to the hypotenuse is half of it: $\displaystyle {MA}={MB}={MC}$ .
So $\displaystyle M$ is at equal distances from $\displaystyle A$ , $\displaystyle B$ and $\displaystyle C$ : it is the centre of the circle through them, and $\displaystyle AB$ , which passes through $\displaystyle M$ , is a diameter.

A use: the circle with the hypotenuse of a right-angled triangle as its diameter passes through the vertex of the right angle.

Exercise 3

$\displaystyle AB$ is a diameter of the circle, and $\displaystyle C$ is on the circle.
$\displaystyle {\angle CAB}={35^{\circ}}$ .

Find

\displaystyle \angle ABC

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle {AB}={10}

and

\displaystyle {AC}={8}

. Find

\displaystyle BC

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

$\displaystyle AB$ is a diameter of the circle with centre $\displaystyle O$ , and $\displaystyle C$ is on the circle.
$\displaystyle CD\perp AB$ , and $\displaystyle D$ is on $\displaystyle AB$ .
Prove: $\displaystyle {\angle ACD}={\angle ABC}$ .

Exercise 5

$\displaystyle A$ , $\displaystyle B$ , $\displaystyle C$ , $\displaystyle D$ are on a circle.
The chords $\displaystyle AC$ and $\displaystyle BD$ meet at $\displaystyle E$ .
Prove: $\displaystyle \triangle AED\sim\triangle BEC$ .

Formulas: geometry, grade 10 (5 units)

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Introduction to analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
The explicit equation of a line – $\displaystyle m$ the slope, $\displaystyle b$ where it crosses the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
The general equation of a line
$\displaystyle Ax+By+C=0$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
A canonical circle – its centre at the origin, $\displaystyle R$ its radius
$\displaystyle x^{2}+y^{2}=R^{2}$
The circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$

Midsegments and the meeting point of the medians

A triangle’s midsegment – $\displaystyle DE$ is parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$
The meeting point of the medians $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Pythagoras’ theorem and its converse: the triangle is right-angled, the right angle opposite $\displaystyle c$
$\displaystyle a^{2}+b^{2}=c^{2}$

Proportion and similarity

A property of a proportion
$\displaystyle \frac{a}{b}=\frac{c}{d}\ \Rightarrow\ \frac{a+b}{b}=\frac{c+d}{d}$
Thales’ theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{DB}$ $\displaystyle =$ $\displaystyle \frac{AE}{EC}$
The extended Thales theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$
The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
Heights, medians, perimeters, the radii of the circumscribed and of the inscribed circles
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{P_1}{P_2}=\frac{R_1}{R_2}=\frac{r_1}{r_2}=k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$

The circle

An inscribed angle is half the central angle on the same arc
$\displaystyle \angle ACB$ $\displaystyle =$ $\displaystyle \frac{1}{2}\angle AOB$
An inscribed angle on the diameter $\displaystyle AB$
$\displaystyle \angle ACB=90^{\circ}$
A quadrilateral $\displaystyle ABCD$ inscribed in a circle – opposite angles
$\displaystyle \angle A+\angle C$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The angle between the tangent $\displaystyle AT$ and the chord $\displaystyle AB$ equals an inscribed angle on $\displaystyle AB$ from its other side
$\displaystyle \angle BAT$ $\displaystyle =$ $\displaystyle \angle ACB$
Two tangents from the point $\displaystyle P$
$\displaystyle PA=PB$
A quadrilateral $\displaystyle ABCD$ circumscribing a circle
$\displaystyle AB+CD$ $\displaystyle =$ $\displaystyle BC+AD$
The chords $\displaystyle AB$ and $\displaystyle CD$ meet at the point $\displaystyle E$
$\displaystyle AE\cdot EB$ $\displaystyle =$ $\displaystyle CE\cdot ED$
Two secants from the point $\displaystyle P$
$\displaystyle PA\cdot PB$ $\displaystyle =$ $\displaystyle PC\cdot PD$
A tangent $\displaystyle PT$ and a secant $\displaystyle PAB$ from the point $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$