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

Answer:

Kindly refer to explanation.

Step-by-step 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.