Theorem 1: Tangents from an external point are equal in length.
Proof:
The angle betwwen the tangent and the radius is 90°. So angle PAO = angle PBO = 90°.
All radii are equal. So if a line is drawn from O to P, the two right-angled triangles that are created will be equal due to the SAS rule.
In short:
angle PAO = angle PBO = 90°
AO = BO
OP is common to both
So length PA = PB
Proof:
The angle betwwen the tangent and the radius is 90°. So angle PAO = angle PBO = 90°.
All radii are equal. So if a line is drawn from O to P, the two right-angled triangles that are created will be equal due to the SAS rule.
In short:
angle PAO = angle PBO = 90°
AO = BO
OP is common to both
So length PA = PB
Theorem 2: The tangent to a circle is perpendicular (90°) to the radius.
Proof:
Draw a chord AC on a circle, and two radii OA and OC at the end of the chord.
Let angle AOC = a
Let angle CAO = b
b = ½(180-a)
As angle a gets closer to 0°, angle b gets closer to 90° and the chord AC becomes tangent to the circle (line ABC).
b = ½(180-a) = ½ x 180 = 90°
Proof:
Draw a chord AC on a circle, and two radii OA and OC at the end of the chord.
Let angle AOC = a
Let angle CAO = b
b = ½(180-a)
As angle a gets closer to 0°, angle b gets closer to 90° and the chord AC becomes tangent to the circle (line ABC).
b = ½(180-a) = ½ x 180 = 90°
Theorem 3: The angle at the centre of a circle is twice the angle at the circumference.
Proof:
Join B to O and extend the line to Q.
Let angle SBO = a
Let angle RBO = b
Triangles SBO and RBO are isosceles
SO = RO = BO = radius of the circle
Hence
angle OSB = a
and
angle ORB = b
angle SOQ = 2a (because angle SOB = 180 - 2a = angle SOQ)
and
angle ROQ = 2b (for the same reason as above)
Hence
angle SBR = a + b
and
angle SOR = 2a + 2b = 2(a + b)
Proof:
Join B to O and extend the line to Q.
Let angle SBO = a
Let angle RBO = b
Triangles SBO and RBO are isosceles
SO = RO = BO = radius of the circle
Hence
angle OSB = a
and
angle ORB = b
angle SOQ = 2a (because angle SOB = 180 - 2a = angle SOQ)
and
angle ROQ = 2b (for the same reason as above)
Hence
angle SBR = a + b
and
angle SOR = 2a + 2b = 2(a + b)
Theorem 4: Angles in the same segment are equal.
Proof:
This proof is based on Theorem 3
Join AO to BO
Let angle AOB = a
Angle APB = ½a (the angle at the circumference is half the angle at the center)
Angle AQB = ½a (the angle at the circumference is half the angle at the center)
∴ angle APB = AQB = ½a
Proof:
This proof is based on Theorem 3
Join AO to BO
Let angle AOB = a
Angle APB = ½a (the angle at the circumference is half the angle at the center)
Angle AQB = ½a (the angle at the circumference is half the angle at the center)
∴ angle APB = AQB = ½a
Theorem 5: Opposite angles of a cylci quadrilateral add to 180°
Proof:
Let angle DAB = a
Let angle DCB = c
Obtuse angle DOB = 2a (the angle at the circumference is half the angle at the center)
Reflex angle DOB = 2c (same reason as above)
Obtuse angle DOB + Reflex angle DOB = 360° (angles around at point add to 360°)
Hence 2a + 2c = 360°
2 (a + c) = 360°
∴ a + c = 180°
Proof:
Let angle DAB = a
Let angle DCB = c
Obtuse angle DOB = 2a (the angle at the circumference is half the angle at the center)
Reflex angle DOB = 2c (same reason as above)
Obtuse angle DOB + Reflex angle DOB = 360° (angles around at point add to 360°)
Hence 2a + 2c = 360°
2 (a + c) = 360°
∴ a + c = 180°
Theorem 6: Angles in a semicircle are 90°
Proof:
Let angle AOB = 180°
Angle APB = ½180° (the angle at the circumference is half the angle at the centre)
∴ angle APB = 90°
Proof:
Let angle AOB = 180°
Angle APB = ½180° (the angle at the circumference is half the angle at the centre)
∴ angle APB = 90°
Theorem 7: Alternate segment theorem.
Proof:
Let BCA = p (p = 60°)
Proof:
Let BCA = p (p = 60°)
Theorem 8: Two radii creat an isosceles triangle.
Proof:
Proof: