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Chapter 7: Circle geo
Geometric Proofs
- Focus on Euclidean Geometry, i.e. angles in a triangle add to $180 \degree$, parallel lines don't intersect, etc.
- There are many kinds of (generally more abstract) geometries, e.g. spherical, projective, hyperbolic, affine, fractal, finite, etc.
Things you are assumed to know already
- Angle sum theorems for triangles and quadrilaterals
- Properties of intersecting lines (congruence of vertically opposite angles).
- Relationships involving angles in parallel lines with a transversal (corresponding, alternate, interior etc.)
- Isosceles triangle theorem (angles opposite congruent sides in a triangle are themselves congruent).
- Triangle congruence and similarity 'tests' (SSS, SAS, etc.)
Circles
The definition of a circle:
A circle is the set of all points at a given distance $r$ from a given point $O$ called the centre.
We define two arcs on the circle as the lengths of the circle separated by two points on the circle. If they are not diametrically opposite, there will be a minor and major arc.
Similarly, the slices of the circle made by the points are two sectors (can be major and minor sectors).
If you join two lines together across the circle, then you create a chord (the line), and the line separates the circle into two segments (once again, if not diametrically opposite you get major and minor segments)
Subtending: The minor arc $AB$ subtends the angle $AOB$ at the centre.
- The minor arc $AB$ subtends the angle ACB at the 'circumference' - i.e. it doesn't have to be at the centre!
Circle Theorems
- An angle at the centre is twice the angle at the circumference subtended by the same arc.
- Prove that $\angle AOC = 2 \times \angle ABC$
- Let $x\degree + y\degree = \angle ABC$
- The line made by $OB$ gives a triangle of $OBC$, and of $OBA$.
- $OA = OB$ (radii)
- $\therefore \triangle AOB$ is isosceles , $\triangle COB$ is isosceles
- $\angle OAB = \angle OBA = x \degree$ ($\triangle AOB$ is isosceles)
- $\angle OAC = \angle OCA = y \degree$
Theorem:
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In a triangle, an exterior angle is the sum of the 2 remote interior angles.
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$\angle AOD = 2\times x \degree$ (exterior angle is sum of remote interior angles)
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$\angle COD = 2 \times y\degree$ (exterior angle is sum of remote interior angles)
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$\angle AOD + \angle COD = 2\times x\degree + 2 \times y \degree = 2(x\degree + y\degree) = 2(\angle ABC)$, as required
- Angles in a semicircle are right angles.
- Prove that if $\overline{AB}$ is a diameter of a circle, then $\angle ACB= 90\degree$.
- IF $\overline{AB}$ is a diameter, it passes through the centre, point $O$.
- Hence, $\angle AOB = 2 \times \angle ACB$ (An angle at the centre is twice the angle at the circumference subtended by the same arc)
- $\angle AOB = 180 \degree$ (angle of the diameter will always be$180\degree$ (its... a straight line))
- $\therefore \angle ACb = \frac{1}{2} \angle AOB = \frac{1}{2} \times 180\degree = 90\degree$, as required
- Angles in the same segment are equal.
- This is essentially using the first theorem for multiple angle on the circumference, because they all subtend the same arc!
- A cyclic quadrilateral is a quadrilateral whose vertices all lie on the same circle. Opposite angles of a cyclic quadrilateral are supplementary.
