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:
[tex]\angle ACB = 90^\circ[/tex]
Solution:
Let [tex]\angle A = x[/tex] and [tex]\angle B = y[/tex].
Construction: Join point A with centre O.
Considering [tex]\triangle AOC[/tex]:
Side AO = OC because both are the radius.
Angles opposite to equal sides in a triangle are equal.
Therefore, [tex]\angle ACO=\angle A=x[/tex]
Considering [tex]\triangle BOC[/tex]:
Side BO = OC because both are the radius.
Angles opposite to equal sides in a triangle are equal.
Therefore, [tex]\angle BCO=\angle B=y[/tex]
[tex]\angle ACB = \angle ACO + \angle BCO = x+y[/tex] ...... (1)
Now, in [tex]\triangle ABC:[/tex]
Using angle sum property of triangle:
[tex]\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[/tex]
By equation (1):
[tex]\angle ACB = 90^\circ[/tex]
Hence proved.
