E.g.

P= 180-(80/2)

P= 180-40

<P= 140°

**Theorem: Sum of opposite angles of a cyclic quadrilateral =180°.**

A quadrilateral is cyclic if it can be inscribed in a circle, with each vertex touching the circumference.

If a quadrilateral has the property that the sums of opposite angles =180°, it is cyclic.

A quadrilateral is cyclic if it can be inscribed in a circle, with each vertex touching the circumference.

If a quadrilateral has the property that the sums of opposite angles =180°, it is cyclic.

**Central angle/Inscribed angle theorem**

**Angles in the same segment**

**Theorem of Thales**

When the two endpoints of the diameter of a circle are connected to a third point on the circumference of the circle, the inscribed angle formed will always be a right angle

This means any triangle inscribed on the diameter of a semicircle will have a right angle

This can be used to find the centre of a circle

**Chord and Angle Properties of Circles**

**Chord and Angle Properties of Circles**

**Cyclic quadrilaterals**

**Intersecting chords theorem**

Proof 1

Proof 1 – using the inscribed angle theorem

Join two opposite corners to the centre (O).

The two angles at the centre = 360° (2D +2A =360°)

Therefore D+A =360°/2 =180°

An inscribed angle is formed when two chords of a circle intersect at a point on the circumference of the circle.

Measurement of the inscribed angle is half the measurement of the central angle.

Bibliography

Community.tes.co.uk, (2014).

Applications in "real" life

. [online] Available at: http://community.tes.co.uk/tes_mathematics/f/25/t/319382.aspx [Accessed 18 May. 2014].

Cut-the-knot.org, (2014).

Thales' Theorem

. [online] Available at: http://www.cut-the-knot.org/pythagoras/ThalesTheorem.shtml [Accessed 17 May. 2014].

Elwes, R. (2010).

Maths 1001

. 1st ed. London: Quercus Publishing, pp.92-96.

Jones, T. and Jackson, S. (2001).

Rugby and Mathematics

. 1st ed. [ebook] pp.1-6. Available at: http://wesclark.com/rrr/rugby_and_math.pdf [Accessed 18 May. 2014].

Mathopenref.com, (2014).

Central Angle Theorem.

[online] Available at: http://www.mathopenref.com/arccentralangletheorem.html [Accessed 17 May. 2014].

Mathopenref.com, (2014).

Thales theorem

. [online] Available at: http://www.mathopenref.com/thalestheorem.html [Accessed 17 May. 2014].

Mathsisfun.com, (2014).

Circle Theorems

. [online] Available at: http://www.mathsisfun.com/geometry/circle-theorems.html [Accessed 17 May. 2014].

Muljadi, P. (2014).

Regiomontanus' angle maximization problem

. [online] Academia.edu. Available at: http://www.academia.edu/3659994/Regiomontanus_angle_maximization_problem [Accessed 17 May. 2014].

Traxler, J. (2014).

Optimal Viewing Angle

. [online] Jwilson.coe.uga.edu. Available at: http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html [Accessed 18 May. 2014].

Wilson, J. (2014).

Angles in a Circle and Cyclic Quadrilateral

. [online] jwilson.coe.uga.edu. Available at: http://jwilson.coe.uga.edu/EMAT6680Su06/Byrd/Assignment%20Nine/CyclicQuad.pdf [Accessed 17 May. 2014].

Angles in Same Segment of Circle are Equal - ProofWiki. 2014. Angles in Same Segment of Circle are Equal - ProofWiki. [ONLINE] Available at: https://proofwiki.org/wiki/Angles_in_Same_Segment_of_Circle_are_Equal. [Accessed 21 May 2014].

A and C are the “end points” B is the “apex point”

Optimal Viewing Angle

Optimal Viewing Angle

-You are in a room with a TV screen 2.5m off the ground and 1m tall. You want to see the movie at the largest vertical viewing angle from eye level. How far away should your head be horizontally from the screen?

Step 3

Results

Inscribed angles continued

Coordinates of the optimal viewing position are (√8.75, 0)

For optimal viewing, you should sit √8.75m, approximately 2.96m, away.

Example

What is the size of angle POQ?

Step 1

Step 2

Figure 1.1

Figure 1. 2

Figure 1.3

Figure 1.4

Figure 1.0

Figure 2.1

References -figures

Figure 1.0: from http://www.webanswers.com/science/math/draw-the-circle-label-the-parts-of-a-circle-1d72cf

Figure 1.1: from http://www.mathsisfun.com/geometry/circle-theorems.html

Figure 1.2: from http://www.mathsisfun.com/geometry/circle-theorems.html

Figure 1.3: from http://www.mathopenref.com/arccentralangletheorem.html

Figure 1.4: from http://www.mathsisfun.com/geometry/circle-theorems.html

Figure 2.1: from http://www.mathopenref.com/thalestheorem.html

Figure 2.2: from http://www.mathopenref.com/thalestheorem.html

Figure 3.1: from http://www.regentsprep.org/Regents/math/geometry/GP14/CircleSegments.htm

Figure 3.2: from http://www.cut-the-knot.org/proofs/IntersectingChordsTheorem.shtml

Figure 3.3: from http://www.mathopenref.com/chordsintersecting.html

Figure 4.1: from https://proofwiki.org/w/images/4/47/Euclid-III-21.png

Figure 4.2: from https://proofwiki.org/w/images/thumb/7/73/IncsribedAngle.PNG/638px-IncsribedAngle.PNG

Figure 5.1: based off

Maths 1001

pg 95

Figure 5.2: based off

Maths 1001

pg 95

Figure 5.3: from http://jwilson.coe.uga.edu/EMAT6680Su06/Byrd/Assignment%20Nine/CyclicQuad.pdf

Figure 6.1: based off http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html

Figure 6.2: based off http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html

Figure 6.3: based off http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html

Figure 6.4: based off http://jwilson.coe.uga.edu/EMAT6680Fa2013/Traxler/Optimal%20Viewing%20Angle/Optimal%20viewing%20angle.html

Proof

The central angle formed by the diameter line PQ is always 180°.

If you draw a line from P and Q to any other point on the circumference of the circle, the inscribed angle theorem tells us that it must create an angle of 90°.

Finding the Centre

The diameter always passes through the centre of the circle

Two straight lines that meet in a 90° point on the circumference of the circle also intersect the circumference at another point

The place where the lines intersect the circumference of the circle will always be the end points of the diameter

Midway between these points is the centre of the circle

Figure 2.2

280°

= 3 because we want the radius to be 3

Substitute the coordinates into the equation

3 - 2.5 = 0.5

Square both sides of the equation to remove the square root

9 - 0.5^2 = 8.75

Square root both sides of the equation to remove the square

X - 0 = X

Figure 5.1

Figure 5.2

Figure 6.1

Figure 6.2

Figure 6.3

1

2

3

4

5

6

Figure 6.4

**Theorem: When two chords intersect each other inside a circle, the products of their segments are equal.**

When 2 chords intersect, 4 angles are created within the circle

When 2 chords intersect, 4 angles are created within the circle

Intersecting Chords rules:

AE x BE = CE x DE

4 angles created

Theorem Proof

Given: Chords AD & BC Intersecting at P

Prove: (AP)(DP) = (CP)(BP)

Triangle APB and Triangle CPD are similar triangles

Angle BAP = Angle DCP. Since both the angles are suspended by same arc

Angle ABP= Angle CDP. Since both the angles are suspended by same arc

Angle APB= Angle CPD. Since both are vertically opposite angles and vertically opposite angles are equal

Using the similar triangle property.

(AP)/(PC) = (BP)/(PD) = (AB)/(CD)

Using the first proportion,

(AP)/(PC) = (BP)/(PD)

On cross multiplying, we get,

(AP)(DP) = (BP)(CP)

Practical Use

In a circle, let angle BAD and angle BED be angles in the same segment, the major arc.

Let F be the center of the circle, and join BF and FD.

From the Inscribed Angle Theorem:

Angle BFD=2 x Angle BAD

Angle BFD=2 x Angle BED

So Angle BAD= Angle BED.

View ABC as a circle, now ∠BEC be an angle at the center. ∠BAC is an

angle at the circumference of the circle.

Now, these angles have the same arc BC at their base.

AE is joined and drawn straight through to F.

Since EA=EB, then from Isosceles Triangles have Two Equal Angles so we can see that Angle EBA= Angle EAB.

So Angle EBA+Angle EAB=2 x Angle EAB.

But from the because the sum of the angles of a triangle =180 = Two Right Angles, we have that Angle BEF=Angle EBA+Angle EAB.

That is, Angle BEF=2x Angle EAB.

For the same reason, Angle FEC=2 x Angle EAC.

So adding them together, we see that Angle BEC=2 x Angle BAC.

Inscribed Angle Theorem

Figure 4.1

Figure 4.2

Theorem: Two angles that are within the same segment (major/minor) are equal

Figure 3.3

Figure 3.1

Figure 3.2

Proof 2

Proof 2 -using angles in the same segment

Figure 5.3

Angle ACB = Angle ADB

Angle BAC = Angle BDC

Angle CAD = Angle CBD

Angle DBA= Angle DCA