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SLIDE 1
Figure 1 shows an arbitrary quadrilateral, and figure 2 shows a cyclic quadrilateral. What do we know about the sum of the interior angles of both figures?
SLIDE 2
In a quadrilateral in general, the sum of the angles is equal to 360°. In addition, if the quadrilateral can be inscribed in a circle (a cyclic quadrilateral), then the sum of opposite angles is 180°.
Quadrilateral, figure 1
Cyclic quadrilateral, figure 2
SLIDE 3
PQRS is a quadrilateral.
? Prove that PQRS is not a cyclic quadrilateral.
? Find the angle x marked in the diagram.
SLIDE 4
? Interior angles at Q and S sum to , which is not 180°. The sum of opposite angles in a cyclic quadrilateral is 180°. Therefore, PQRS is not a cyclic quadrilateral.
? The interior angle sum of a quadrilateral is 360°. Therefore,
SLIDE 5
Prove that the interior angle sum of a quadrilateral is 360°.
SLIDE 6
Any quadrilateral can be divided into two triangles. The sum of the angles in each triangle is 180°. The sum of the interior angles of the quadrilateral is the sum of the angles of the two triangles. Therefore, the sum of the interior angles of a quadrilateral is 360°.
SLIDE 7
Find x and y and all the individual interior angles.
SLIDE 8
Interior angles B and D sum to 180°.
Interior angles A and C sum to 180°.
SLIDE 9
Consolidation
Prove that the opposite angles of a cyclic quadrilateral add up to 180°.
SLIDE 10
Join each vertex of the quadrilateral to the centre of the circle O.
because they are all radii of the circle.
Each of the triangles are isosceles and angles that are equal are marked onto the diagram.
The angle sum of a cyclic quadrilateral is 360°.
The angle A is , the angle B is , so the sum of opposite angles is 180°.
SLIDE 11
What is the fundamental property of the circle that enables us to prove that opposite angles of a cyclic quadrilateral add up to 180°?
SLIDE 12
Fundamental property of a circle
What makes a circle is a circle is that any point on the circle is the same distance from the centre of the circle.
Any two radii are equal.
That makes any triangle like OAB in the diagram, where A and B lie on the circle (the circumference), into an isosceles triangle.
In any problem that depends on the fact that a geometric object is a circle, draw in the radii.
SLIDE 13
The tangent to any curve is a straight line that just touches it.
The tangent to a circle is perpendicular to a radius.
SLIDE 14
Find the angle .
SLIDE 15
As AB and AC are tangents to the circle, .
Therefore,
OBC is an isosceles triangle, because of the radii, so
SLIDE 16
Circle terminology
In diagram ?, the line AB, where A and B are two points on the circumference of a circle, is called a chord.
In diagram ?, the part of the circle between P and Q and marked s is called an arc.
In diagram ?, angle x is called the angle subtended on the arc TU.
In diagram ?, angle y is called the angle subtended at the circle.
SLIDE 17
Circle theorem – angle at centre
In any circle, for any arc AB, the angle subtended at the centre is twice the angle subtended on the arc.
In the diagram, .
Task
Prove the theorem.
Hint. Join the radius OX and use the fundamental property of a circle to find isosceles triangles.
SLIDE 18
Theorem
In any circle, for any arc AB, the angle subtended at the centre is twice the angle subtended on the arc.
Proof
Join O to X. Then are all radii, and AOX, BOX are isosceles triangles.
Angles a are equal base angles of the isosceles triangle AOX. , so . That is b is twice the angle a.
By a similar argument, . Since and , then
Proven
SLIDE 19
Formal proof
We have proven a lot of things in this course in the past. Now is the time to introduce the idea of a formal proof.
In a formal proof the proposition (a statement) that is to be proven is stated at the beginning.
Then there follows the proof, which is a series of rigorous steps working from known principles to the conclusion, which is the theorem.
At the end, we write something to show that the proof is concluded.
Options include the word, “proven”, the expression “QED”, which is Latin for quot erat demonstrandum, or in English, “that which was to be demonstrated”, and a symbol such as a solid square, ?.
SLIDE 20
Find the angle y.
SLIDE 21
By the circle theorem (angle at centre)
SLIDE 22
Find the angle .
SLIDE 23
The greater angle at the centre is .
This is twice the angle subtended on the arc, so
AXBO is a quadrilateral,
SLIDE 24
Circle theorem – angle in a semicircle
If AOB is a diameter of a circle, then the angle subtended on the arc AXB, which is a semicircle, is a right angle.
Prove the theorem.
Hint. You may assume the first circle theorem on the angle at the centre.
SLIDE 25
Theorem
If AOB is a diameter of a circle, then the angle subtended on the arc AXB, which is a semicircle, is a right angle.
Proof
The first theorem states: in any circle, for any arc, the angle subtended at the centre is twice the angle subtended on the arc.
Here the angle subtended at the centre is 180°. The angle subtended on the arc is therefore half of this, 90°. ?
(This is only a special case of the first theorem.)
SLIDE 26
AOB is a diameter. Find the angles and
SLIDE 27
is a right angle, by the theorem.
Hence, .
OCB is isosceles.
Hence, .
SLIDE 28
AOB is a diameter.
The ratio of angle x to angle y is
Find the size of angle z.
SLIDE 29
is a right angle, by the theorem. Hence,
Therefore, .
Hence, .
SLIDE 30
Find the angle x.
SLIDE 31
AD is a tangent. is a right angle.
Therefore, .
.
ODC is isosceles.
.
SLIDE 32
Circle theorem – angles in the same segment
Two angles subtended on the same segment of an arc in a circle are equal.
Prove this theorem.
Hint. Like the angle in semicircle theorem, it is a direct application of the first theorem – the angle at centre.
SLIDE 33
Theorem
Two angles subtended on the same segment of an arc in a circle are equal.
Proof
The first theorem states: in any circle, for any arc, the angle subtended at the centre is twice the angle subtended on the arc.
Construct the angle subtended on the centre of the circle. Then both angels subtended in the same segment of the arc AB are half this angle, by the first theorem. Hence, they are equal. ?
SLIDE 34
Prove that the triangles APB and CPD are similar.
SLIDE 35
The angles marked x are in the same segment of the arc ABDC, and the angles marked y are in the same segment of the arc BACD. The angles marked y are vertically opposite. Both triangles have corresponding angles, and are therefore similar.
SLIDE 36
BPOC is a diagonal. All measurements are in cm. Find the radius of the circle, giving your answer as a fraction.
SLIDE 37
PAB and PCD are similar.
SLIDE 36B
BPOC is a diagonal. All measurements are in inches. Find the radius of the circle, giving your answer as a fraction.
SLIDE 37B
PAB and PCD are similar.
SLIDE 38
AB is a diagonal.
All measurements are in cm. Find the length DB. Give your answer to 3 significant figures.
SLIDE 39
Since AB is a diameter, the triangles are both right-angled triangles.
We are given: .
ACB is a right-angled triangle with sides in the Pythagorean ratio, 3 to 4 to 5. That is, .
Then, in the triangle ADB
SLIDE 38B
AB is a diagonal.
All measurements are in inches. Find the length DB. Give your answer to 3 significant figures.
SLIDE 39B
Since AB is a diameter, the triangles are both right-angled triangles.
We are given: .
ACB is a right-angled triangle with sides in the Pythagorean ratio, 3 to 4 to 5. That is, .
Then, in the triangle ADB
SLIDE 40
Circle theorem – alternate segment theorem
PQ is tangent to the circle at A. Then, angles in alternate segments are equal.
This means the angles marked x and y are equal.
Prove the theorem.
Hint. You need to use the property of a tangent to a circle, the fundamental property of circle and the first theorem on angle at the centre.
SLIDE 41
Theorem
PQ is tangent to the circle at A. Then, angles in alternate segments are equal.
Proof
Let the angle . We aim to show that .
Construct the line AO. As PQ is tangent, this is a radius, and O is the centre of the circle. Then by the angle at centre theorem, the angle is twice the angle subtended at the arc segment; that is, .
Triangle AOC is isosceles, so . This gives, . But . Hence, . ?
SLIDE 42
What is the size of the angle marked x?
SLIDE 43
By the alternate segment theorem, .
The diagram on the right shows you how to remember which segment is alternate.
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