# Circle Theorems#

[IMAGE OF ALL CIRCLE THEOREMS]

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Proof:

Let angle DBA = Y and angle DBC = X

Image 1

In the image above we can see an arrow shape which can be separated into two isosceles triangles (triangle ABD, and triangle BCD). We know they are isosceles triangles because sides DA, DB and DC are all radii so have the same length.

Now we have established the two triangles are isosceles, angle DBC(x) = angle DAB and angle DBA(Y) = angle DCB

This is true because the isosceles triangle theorem states an isosceles triangle must have two equal length sides and that the angles opposite to each of the equal sides must also be equal

Image 2

Angle BDA = 180 - (x + x) Or Angle BDA = 180 - 2x (because angles in a triangle add up to 180°)

Angle BDC = 180 - (y + y) Or Angle BDC = 180 - 2y (because angles in a triangle add up to 180°)

Image 3

Image 1

Proof:

The angle AOC = 180° (because the angle on a straight line is 180°)

Image 2

Due to circle theorem 1 which states:

The angle at the centre is twice the angle at the circumference

Angle x = 1/2 of angle AOC Angle x = 1/2 x 180 Angle x = 90°

Image

Proof:

Image1

Because of circle theorem 1 which states:

The angle at the centre is twice the angle at the circumference

This means that: angle ABD = 2x / 2 angle ABD = x

and

angle ACD = 2x / 2 angle ACD = x

Image 2

Since angle ACD = x and angle ABC = x

angle ACD = angle ABC

A cyclic quadrilateral is a quadrilateral drawn inside a circle. Important To be a cyclic quadrilateral, every corner of the quadrilateral must touch the circumference of the circle.

This means angle between the tangent and the chord at the point of contact is equal to the angle in the alternate segment