A semi circle is the half of the circle. The circle is divided in exactly two halves by any diameter of the circle. The diameter is also the longest chord that passes through the center and divides the circle in two equal sectors. Also, the diameter is exactly double the radius as it passes through the center.

The angle made by two points lying on the circle at any other point on the circle is known as the inscribed angle. In this article, we are going to understand about the concept of angle is a semi circle.

According to the angle in a semi circle theorem, the angle inscribed in a semi circle is always a right angle.

We can prove this theorem very easily. We know that the angle inscribed in the circle is half of the measure of the angle made at the centre by the same points. Now, angle in a semi circle means, the points making that angle are the end points of the diameter of the circle. Now angle made by the diameter at the centre of the circle measure $180^{\circ}$, since, it passes through the centre. Therefore, the angle made by the diameter at any other point on the circle measures half of $180^{\circ}$, that is, $90^{\circ}$, which is a right angle.

Therefore, the angle in a semi circle will always be a right angle.

The triangle so formed by the diameter and the arms of the angle in a semi circle will thus always be a right angled triangle. So we can also say that any triangle inscribed in the semi circle with one side being the diameter of the circle will always be a right-angled triangle. Also, conversely, an inscribed triangle in the circle is a right angled triangle then the triangle lies in the semi circle with diameter as one of its side.

Another method to prove this is below.

Semi Circle Theorem

Angle ACB is an angle in a semi circle. We need to show that it is a right angle.

Let angle CAB = p and angle CBA = q

We draw a line from the centre of the circle to the point C.

Since AD and DC are radii of the circle, therefore angle CAD = angle DCA = p

Similarly, angle CBD = angle DCB = q

We know that the sum of the angles of a triangle is 180?.

Therefore, $p\ +\ q\ +\ (p + q)\ =\ 180$

$\Rightarrow\ 2\ (p\ +\ q)\ =\ 180$

$\Rightarrow\ p\ +\ q$ = $\frac{180}{2}$ = $90^{\circ}$

$\Rightarrow\ angle\ ACB\ =\ 90^{\circ}$

Therefore, angle ACB is a right angle. Hence, the angle in a semi circle is a right angle.
It is to be noted that since any angle in a semi circle is a right angle so it measures $90^{\circ}$ exactly. So, when we are given that certain angle lies in a semi circle then it is taken as $90^{\circ}$ only and if we are given that the inscribed angle in a circle measures $90^{\circ}$, this means that angle lies in the semi circle.
When we require drawing an angle in a semi circle when we are given the vertex of the angle only, we can follow the steps below.

1) First we mark a point on the circle that will be the vertex of the angle.

2) Now we draw the any diameter of that circle.

3) We now join the end points of the diameter to that point to form the angle in a semi circle.
When we are given the length of an arm of the angle, then we follow the steps below to draw the angle in a semi circle.

1) We plot a point on the circle.

2) With that point and radius equal to the measure of the length of the arm of the angle we make an arc intersecting the circle in another point.     Join the two points and measure to verify.

3) Now with the first point of step 1, we draw a line passing through the center of the circle intersecting the circle in another point.

4) We join the point of step 2 with the point obtained in step 3.

5) Now, the angle so formed will be the angle in a semi circle.
Let us see some examples on the basis of angle in a semi circle concept.
Example 1: 

Find x in the figure below
 
Semi Circle Example

Solution:

We know that angle in a semi circle is a right angle

This implies that $2x\ +\ 6\ =\ 90$

$\Rightarrow\ 2x\ =\ 90\ - \ 6\ =\ 84$

$\Rightarrow\ x$ = $\frac{84}{2}$ = $42$
Example 2: 

In the given figure, find x.

Semi Circle Examples
 
Solution: 

The angle in a semi circle is a right angle. 

Therefore, angle $ABC\ =\ 90^{\circ}$

$\Rightarrow\ 90\ +\ 63\ +\ x\ =\ 180$

(the sum of angles of a triangle is $180^{\circ}$)

$\Rightarrow\ x\ +\ 153\ =\ 180$

$\Rightarrow\ x\ =\ 180\ -\ 153$

$\Rightarrow\ x\ =\ 27^{\circ}$