Circle Theorems: Angles in Semicircles

Definition of Angles in Semicircles

An angle in a semicircle is the angle formed at any point on the circumference of a circle when the endpoints of the diameter are joined to that point. The key property is:

• The angle subtended by the diameter of a circle at the circumference is always a right angle (i.e., $90^\circ$).

This means if you draw a triangle where one side is the diameter of the circle, the angle opposite that side (at the circumference) will always be $90^\circ$.

Basic terms:

• Circle: The set of all points equidistant from a fixed point called the centre.
• Diameter: A straight line passing through the centre of the circle with endpoints on the circle.
• Semicircle: Half of a circle, formed by the diameter and the arc connecting its endpoints.

This theorem is sometimes called the Thales’ theorem and is fundamental in circle geometry.

For instance, if a circle has a diameter $AB$, and $C$ is any point on the circle (but not on the line $AB$), then the angle $ACB$ is a right angle:

$\angle ACB = 90^\circ$

Example: If diameter $AB = 8\,\mathrm{cm}$, then angle $ACB = 90^\circ$ by the theorem.

Proof of the Theorem

The proof uses properties of triangles and isosceles triangles within the circle.

Consider a circle with centre $O$, diameter $AB$, and a point $C$ on the circumference. Join $O$ to $C$, and also join $A$ to $C$ and $B$ to $C$.

• Since $OA$, $OB$, and $OC$ are all radii of the circle, they are equal in length: $OA = OB = OC$.
• Triangles $OAC$ and $OBC$ are both isosceles, because they each have two sides equal (radii).

Let the angle at $O$ between $OA$ and $OC$ be $\theta$, and the angle at $O$ between $OB$ and $OC$ be $180^\circ - \theta$ because $A$, $O$, and $B$ lie on a straight line (diameter).

Using the properties of isosceles triangles:

• In $\triangle OAC$, the base angles at $A$ and $C$ are equal. Let each be $\alpha$.
• In $\triangle OBC$, the base angles at $B$ and $C$ are equal. Let each be $\beta$.

The angle at $O$ in $\triangle OAC$ is $\theta$, so the sum of angles in $\triangle OAC$ gives:

$\theta + 2\alpha = 180^\circ \Rightarrow 2\alpha = 180^\circ - \theta$

Similarly, in $\triangle OBC$, the angle at $O$ is $180^\circ - \theta$, so:

$(180^\circ - \theta) + 2\beta = 180^\circ \Rightarrow 2\beta = \theta$

The angle at $C$ in the larger triangle $ABC$ is $\alpha + \beta$. Using the above, sum these:

$\alpha + \beta = \frac{180^\circ - \theta}{2} + \frac{\theta}{2} = \frac{180^\circ}{2} = 90^\circ$

Thus, $\angle ACB = 90^\circ$, proving the angle in a semicircle is a right angle.

Applications and Examples

This theorem is very useful for finding unknown angles in circle problems, especially when a triangle is formed using the diameter as one side.

For example, if you know two points on a circle form a diameter, any triangle formed with these points and a third point on the circle must have a right angle opposite the diameter.

This can be applied to solve problems involving:

• Finding unknown angles in triangles inscribed in circles.
• Proving triangles are right angled without measuring angles directly.
• Solving geometric problems involving semicircles and chords.

Although cyclic quadrilaterals involve other circle theorems, the angle in a semicircle theorem can sometimes be used as a step in solving problems involving cyclic shapes.

For instance, if a triangle is inscribed in a circle with one side as the diameter, you can immediately conclude the triangle is right angled, which can simplify calculations or proofs.

Example: Suppose a circle has diameter $AB = 10\,\mathrm{cm}$. Point $C$ lies on the circle forming triangle $ABC$. Find $\angle ACB$.

Since $AB$ is the diameter, the angle at $C$ is $90^\circ$ by the theorem.