Circle Theorems and other geometry

ray.c.carpenter, Mar 23, 2016

Table of Contents

  1. Circle Theorems
    1. Angle at centre and circumference
    2. Angle in semicircle
    3. Angles in the same segment
    4. Opposite angles in a cyclic quadrilateral
    5. Tangents from a point
    6. Angle between tangent and radius
    7. Alternate segment
    8. Perpendicular from centre to chord
  2. van Schooten's theorem
    1. van Schooten's theorem start point
    2. van Schooten's theorem extended diagram for proof
  3. Pythagoras drawing for proof
    1. Static Pythagoras drawing for proof
    2. Dynamic Pythagoras drawing for proof
  4. Paper Folding
    1. Equilateral triangle from a rectangle - diagram 1a
    2. Equilateral triangle from a rectangle - diagram 1b
    3. Alternative equilateral triangle diagram 2a
    4. Alternative equilateral triangle diagram 2b
    5. Equilateral triangle from a circle

1.

Circle Theorems

  • 1. Angle at centre and circumference
  • 2. Angle in semicircle
  • 3. Angles in the same segment
  • 4. Opposite angles in a cyclic quadrilateral
  • 5. Tangents from a point
  • 6. Angle between tangent and radius
  • 7. Alternate segment
  • 8. Perpendicular from centre to chord

1.1.

Angle at centre and circumference

The angle at the centre is twice the angle at the circumference.
– ray.c.carpenter

1.2.

–

1.3.

–

1.4.

–

1.5.

–

1.6.

–

1.7.

–

1.8.

–

2.

van Schooten's theorem

  • 1. van Schooten's theorem start point
  • 2. van Schooten's theorem extended diagram for proof

2.1.

van Schooten's theorem start point

[url=http://www.cut-the-knot.org/Curriculum/Geometry/Pompeiu.shtml#explanation]van Schooten's Theorem[/url] asserts that, for a point P on the circumcircle of an equilateral triangle ABC, the length of the longest of the segments (PA, PB, PC) is the sum of the shorter two.
– ray.c.carpenter

2.2.

–

3.

Pythagoras drawing for proof

  • 1. Static Pythagoras drawing for proof
  • 2. Dynamic Pythagoras drawing for proof

3.1.

Static Pythagoras drawing for proof

Four identical right angled triangles have been arranged so that they form an outer square and also have an inner square made from the hypotenuses. The area of the outer square equals the area of the inner square plus the area of the 4 triangles [br]ie (a+b)² =(4 x ab/2) +c², which simplifies to a² + b² = c²
– ray.c.carpenter

3.2.

–

4.

Paper Folding

  • 1. Equilateral triangle from a rectangle - diagram 1a
  • 2. Equilateral triangle from a rectangle - diagram 1b
  • 3. Alternative equilateral triangle diagram 2a
  • 4. Alternative equilateral triangle diagram 2b
  • 5. Equilateral triangle from a circle

4.1.

Equilateral triangle from a rectangle - diagram 1a

– ray.c.carpenter

4.2.

–

4.3.

–

4.4.

–

4.5.

–

Information