Angle in a semicircle is a right angle

Another one of popular and important theorems in circles is the one which states that the angle in a semicircle is a right angle. 

                                 OR

The angle subtended by a diameter at any point on the circumference of a circle is a right angle. Both are just two different ways of stating the same thing. 

This theorem is actually a direct outcome of another theorem on circles which was also the topic of my previous post. So let’s quickly review the theorem first that I wrote about in my last post. To avoid confusing my brain, I am gonna call it Theorem 1. And the theorem whose statement is mentioned above as Theorem 2, which I will show you is a special case of Theorem 1. 

Let’s quickly get familiar with some basic terms. An arc is any part of the circumference(boundary) of a circle. In the given figure, AFB is an arc of the circle with center O. It is a minor arc because it is smaller than half the length of the circumference of the circle. 

Let P be a point anywhere on the circumference(excluding the arc). Connecting P with the endpoints of arc AFB results in the formation of angle APB. This angle is the angle subtended by the minor arc AFB at point P on the (rest of the) circumference of the circle.

Also, joining center O with endpoints A and B. The resulting angle AOB is the angle subtended by the minor arc AFB at the center of the circle.  

Theorem 1 states that the angle subtended by an arc at any point on the (rest of the) circumference of a circle is half the angle subtended by that arc at the center. 

So if x is the angle subtended by minor arc AFB at center O, \(\frac{x}{2}\) will be the angle it will subtend at (any) point P on the circumference. 

If ∠AOB = x, 

⇒∠APB = \(\frac{x}{2}\)

Here AFB is a minor arc, but the result holds true for angles subtended by a major arc as well. A major arc is bigger than half the circumference of a circle. 

A point to note here is that a minor arc will always subtend an angle less than 180° at the center, and a major arc will always subtend an angle greater than 180° at the center. But what if the length of an arc was exactly equal to half the circumference of the circle? 

In this case, the angle subtended by an arc at the center is exactly 180°. A single line segment(diameter) is connecting the endpoints of the arc and the center of the circle as we see in the figure, which was not the case in the previous two situations. 

Theorem 1 is valid for all kinds of arcs, so it is valid in this case as well. The angle subtended by arc AFB at (any) point P on the (rest of the) circumference is \(\frac{180°}{2} = 90°\). 

The segment AB is a diameter which bisects the circle into two equal regions or semicircles. And so P being a point on the boundary of a semicircle, angle APB is said to be the angle in a semicircle.  

And so hopefully now the statement of Theorem 2, ‘the angle in a semicircle is a right angle’ makes sense. We can see that Theorem 2 is just a special case of Theorem 1. 

Angle APB is subtended by arc AFB. Or we can also say it is the angle subtended by diameter AB. Hence the statement, the angle subtended by a diameter at any point on the circumference is a right angle. 

I will present another neat little way to prove theorem 2. Taking the same figure, we will prove that Angle APB is 90° by using properties of an isosceles triangle. First we will draw a segment from A to B if we haven’t done so before.

Connecting the center O to P with a straight segment will result in the formation of two isosceles triangles. Triangle OAP and triangle OBP. An isosceles triangle is a triangle with two equal sides. 

In triangle OAP, 

OA = OP (radii)

∴∠OAP = ∠OPA = x (say) (opposite angles to equal sides are also equal) 

Angle POB is an exterior angle of triangle OAP. If you extend any side of a triangle, the angle that the extended portion makes with the side adjacent to the side that you extended is an exterior angle of that triangle. 

In the above figure, OB is an extended portion of side OA of triangle OAP. And angle POB is the angle it makes with the adjacent side OP. 

As per the exterior angle theorem, the external angle is equal to the sum of opposite interior angles. 

∠POB = ∠OAP + ∠OPA

                  = x  +  x

                  = 2x

In triangle OBP, 

OB = OP (radii)

∴∠OBP = ∠OPB = y (say) (opposite angles to equal sides are also equal)

The three interior angles of triangle OBP sum up to 180°. 

y + y + 2x = 180°

2y = 180° - 2x

y = \(\frac{180° - 2x}{2}\) 

 

  = 90° - x 

We want to prove ∠APB =  90°.

From the figure above,

∠APB = x + y

           = x + 90° - x

           = 90°

Hence proved.