A triangle inscribed in a circle

Geometry, grade 11 (4 units) Lesson 10 of 20Learning

A triangle inscribed in a circle

A triangle is inscribed in a circle when its three vertices are on the circle. That circle is the triangle’s circumscribed circle.

Theorem: Every triangle can be inscribed in a circle.

The centre of the circumscribed circle is equally far from the three vertices – the radius. So we look for such a point.

The meeting point of the perpendicular bisectors

From grade 10: the perpendicular bisector of a segment is the locus of all the points equally far from its ends. That is, every point on it is equally far from the ends, and every point equally far from the ends is on it.

Theorem: In a triangle, the three perpendicular bisectors meet at one point, the centre of the circumscribed circle.

Given: $\displaystyle M$ is the midpoint of $\displaystyle AB$ , and $\displaystyle N$ the midpoint of $\displaystyle BC$ .
The perpendicular bisectors $\displaystyle SM\perp AB$ and $\displaystyle SN\perp BC$ meet at $\displaystyle S$ .
To prove: $\displaystyle S$ is on the perpendicular bisector of $\displaystyle AC$ too, and $\displaystyle {SA}={SB}={SC}$ .

The proof below shows that $\displaystyle S$ is equally far from the three vertices. Both things follow from that: the third perpendicular bisector passes through $\displaystyle S$ , and the circle with centre $\displaystyle S$ passes through $\displaystyle A$ , $\displaystyle B$ and $\displaystyle C$ .

	Claim	Reason
1	$\displaystyle {SA}={SB}$	$\displaystyle S$ is on the perpendicular bisector of $\displaystyle AB$ : every point on a segment’s perpendicular bisector is equally far from its ends
2	$\displaystyle {SB}={SC}$	$\displaystyle S$ is on the perpendicular bisector of $\displaystyle BC$ , for the same reason
3	$\displaystyle {SA}={SC}$	(1), (2)
4	$\displaystyle SK\perp AC$	(3); $\displaystyle K$ is the midpoint of $\displaystyle AC$ . A point equally far from a segment’s ends is on its perpendicular bisector – $\displaystyle SK$ is the perpendicular bisector of $\displaystyle AC$
5	$\displaystyle {SA}={SB}={SC}$	(1), (2); so the circle with centre $\displaystyle S$ and radius $\displaystyle SA$ passes through the three vertices

Where the centre of the circumscribed circle is

Where the centre is depends on the triangle’s largest angle:

in an acute-angled triangle – inside the triangle;
in a right-angled triangle – at the midpoint of the hypotenuse (the lesson on an inscribed angle on a diameter);
in an obtuse-angled triangle – outside the triangle, beyond the side opposite the obtuse angle.

Example: a perpendicular bisector in the coordinate system

The vertices of the triangle are $\displaystyle A(6,8)$ , $\displaystyle B(7,1)$ and $\displaystyle C(-2,4)$ .
We find the equation of the perpendicular bisector of the side $\displaystyle BC$ .

The midpoint of $\displaystyle BC$ : $\displaystyle {\left(\frac{7+(-2)}{2},\frac{1+4}{2}\right)}={(2.5,2.5)}$ .
The slope of $\displaystyle BC$ : $\displaystyle {m_{BC}}={\frac{4-1}{-2-7}}={-\frac{1}{3}}$ .
The bisector is perpendicular to $\displaystyle BC$ , so its slope is $\displaystyle 3$ (the product of the slopes is $\displaystyle -1$ ).
The line of slope $\displaystyle 3$ through $\displaystyle (2.5,2.5)$ :

y-2.5=3(x-2.5)\quad\Rightarrow\quad y=3x-5

Exercise 1

We go on with the triangle of the example: $\displaystyle A(6,8)$ , $\displaystyle B(7,1)$ , $\displaystyle C(-2,4)$ .
The perpendicular bisector of $\displaystyle BC$ is $\displaystyle {y}={3x-5}$ .

Find the equation of the perpendicular bisector of the side $\displaystyle AB$ , and write it in a row of the graph board.

\displaystyle A=(6,8)

\displaystyle B=(7,1)

\displaystyle C=(-2,4)

\displaystyle y=3x-5

Find the point

\displaystyle S

where the two perpendicular bisectors meet.

Find

\displaystyle SA

. (Find

\displaystyle SB

and

\displaystyle SC

too – what do you get?)

/ – 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

Write the equation of the circle circumscribed about the triangle $\displaystyle ABC$ .

\displaystyle A=(6,8)

\displaystyle B=(7,1)

\displaystyle C=(-2,4)

Exercise 2

The vertices of the triangle are $\displaystyle A(5,6)$ , $\displaystyle B(-1,2)$ and $\displaystyle C(0,7)$ .

Find the product of the slopes of

\displaystyle AC

and

\displaystyle BC

. What kind of triangle is

\displaystyle 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

The centre of the circumscribed circle is the midpoint of the hypotenuse

\displaystyle AB

. Find the radius.

/ – 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

Write the equation of the circle circumscribed about the triangle $\displaystyle ABC$ .

\displaystyle A=(5,6)

\displaystyle B=(-1,2)

\displaystyle C=(0,7)

Exercise 3

$\displaystyle S$ is the centre of the circle circumscribed about the triangle $\displaystyle ABC$ .
Prove: $\displaystyle {\angle SAB}={\angle SBA}$ .

Formulas: geometry, grade 11 (4 units)

Open as a page

The equation of a circle

The canonical equation – the centre at the origin, the radius $\displaystyle R$
$\displaystyle x^{2}+y^{2}=R^{2}$
The general equation – the centre $\displaystyle (a,b)$ , the radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$
The distance between two points
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of a segment
$\displaystyle x_M=\frac{x_1+x_2}{2},\ y_M=\frac{y_1+y_2}{2}$
Perpendicular lines – $\displaystyle m_1$ , $\displaystyle m_2$ their slopes
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$

Angles and chords in a circle

An inscribed angle – half the central angle on the same arc
$\displaystyle \angle ACB$ $\displaystyle =$ $\displaystyle \frac{1}{2}\angle AOB$
A quadrilateral inscribed in a circle – a pair of opposite angles
$\displaystyle \angle A+\angle C$ $\displaystyle =$ $\displaystyle 180^{\circ}$
A chord $\displaystyle AB$ – $\displaystyle d$ its distance from the centre
$\displaystyle \left(\frac{AB}{2}\right)^{2}+d^{2}$ $\displaystyle =$ $\displaystyle R^{2}$

Tangents

Two tangents from a point $\displaystyle P$ – $\displaystyle A$ , $\displaystyle B$ the points of contact
$\displaystyle PA=PB$
The tangent segment $\displaystyle PA$ – $\displaystyle O$ the centre
$\displaystyle PA^{2}+R^{2}$ $\displaystyle =$ $\displaystyle PO^{2}$

Trigonometry, circumference and area

The law of sines – $\displaystyle R$ the radius of the circle through the triangle’s vertices
$\displaystyle \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R$
The area of a triangle – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$
The area of a parallelogram – $\displaystyle a$ , $\displaystyle b$ adjacent sides, $\displaystyle \alpha$ the angle between them
$\displaystyle A$ $\displaystyle =$ $\displaystyle ab\sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The circumference
$\displaystyle C$ $\displaystyle =$ $\displaystyle 2\pi R$
The area of a disc
$\displaystyle A$ $\displaystyle =$ $\displaystyle \pi R^{2}$