Question

proof that angle subtended at center is twice at any other pointon circle

Answer 1

Given: Arc AB
            Point C on the circle is outside AB.

To prove: ∠AOB = 2 × ∠ACB

Construction: Draw a line CO extended till point D.

Proof: In ΔOAC in each of these figures,
            ∠AOD = ∠OAC + ∠OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles)

                 OA = OC (Radii of same circle)
Thus,  ∠ OAC = ∠ OCA (Angles opposite equal sides of a triangle are equal)
            ∠AOD = ∠OAC + ∠OCA
         ⇒∠AOD = 2 × ∠OCA

Similarly, in ΔOBC,
∠BOD = 2 × ∠OCB
∠AOD = 2 × ∠OCA and ∠BOD = 2 × ∠OCB
⇒ ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB
∠AOD + ∠BOD = 2 × (∠OCA + ∠OCB)
or ∠AOB = 2 × ∠ACB

Hence, the theorem is proved.

Answer 2

(a) - Firstly, you should draw a circle and mark its centre. Once you have done this, draw a diameter through the centre. Use this diameter to form one side of a triangle. The other two sides should meet along the circumference of the circle. Your diagram should look like this:

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Secondly, you need to divide the triangle in two by drawing a radius from the centre to the vertex on the circumference:

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By doing so, you can recognise that these smaller triangles have two identical sides which are both the length of the radius of the circle. Due to the fact that each of these small triangles have two sides of equal length, they must both be isosceles triangles. Subsequently, according to the rules of isosceles triangles, both of these small triangles must have equal angles.

According to the rules of triangles, all of the angles within a triangle must add up to 180 °. By observing the angles of the two small triangles, it is evident that the three angles within the large triangle are: a, b, and a + b. Using these symbols, you can form the equation:

2a+2b=180
2a+2b=180
This can be written as:

a+b=90
a+b=90
Therefore, you have proved that the angle which is subtended at the circumference by a semicircle adds up to 90°.

2 - How to prove that the angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference
In order to prove that the angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference, you must follow this process:

Example
(a) - Prove that the angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference

Solution
(a) - Firstly, draw a circle and mark its centre. Once you have done this, choose any two points on the circumference below the centre and one point on the circumference above the centre. Draw a line which connects each point below the centre to the centre itself and to the point on the circumference above the centre. Label the angle at the centre 'a' and the angle at the circumference 'b'. You diagram should look like this:

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Now you need to prove that a = 2b.

In order to do this, you need to draw a line from the centre to the point on the circumference above the centre. This line is the radius to the circle. The other two lines from the centre to the circumference are also radii to the circle. Due to the fact that all radii in a circle are equal, the two triangles which you have formed must be isosceles triangles (due to the fact that they have equal side lengths):

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Now that you have proved that these two small triangles are both isosceles, they must both have 2 equal angles. If you combine this knowledge with the fact that all the angles within a triangle add up to 180⁰, you can form the equations:

2w+x=180
2w+x=180
2y+z=180
2y+z=180
Therefore, you can re-arrange these equations to demonstrate that:

x=180−2w
x=180−2w
z=180−2y
z=180−2y
Given that the angles which surround a point add up to 360⁰, you can form the equation:

a+x+z=360
a+x+z=360
a+(180−2w)+(180−2y)=360
a+(180−2w)+(180−2y)=360
This equation can be rearranged such that:

a−2w−2y=0
a−2w−2y=0
a=2(w+y)
a=2(w+y)
Given that b = w + y, you can conclude that a = 2b

Therefore, you have proved that the angle which is subtended by an arc at the centre of a circle is double the size of the angle subtended at any point on the circumference