Question

A, B, and C are points on the circumference of a circle, centre 0. AOB is a diameter of the circle. Prove that angle ACB is 90° You must not use any circle theorems in your proof.

Answer 1

Kindly refer to explanation.

Explanation:

Given that:

A,B and C are 3 points on circumference of circle with centre O.

AOB is diameter.

Kindly refer to the attached image.

To prove:

`\angle ACB = 90^\circ`

Solution:

Let
`\angle A = x` and
`\angle B = y`.

Construction: Join point A with centre O.

Considering
`\triangle AOC`:

Side AO = OC because both are the radius.

Angles opposite to equal sides in a triangle are equal.

Therefore,
`\angle ACO=\angle A=x`

Considering
`\triangle BOC`:

Side BO = OC because both are the radius.

Angles opposite to equal sides in a triangle are equal.

Therefore,
`\angle BCO=\angle B=y`

`\angle ACB = \angle ACO + \angle BCO = x+y` ...... (1)

Now, in
`\triangle ABC:`

Using angle sum property of triangle:

`\angle A + \angle B + \angle ACB =180^\circ\\\Rightarrow x+y+x+y=180^\circ\\\Rightarrow 2(x+y)=180^\circ\\\Rightarrow x+y=90^\circ`

By equation (1):

`\angle ACB = 90^\circ`

Hence proved.