Angles in Semicircles

The angle in a semicircle theorem

Let $AB$ be a diameter of a circle and let $C$ be any other point on its circumference. Joining $A$ and $B$ to $C$ forms triangle $ABC$. The angle opposite the diameter is always a right angle:

$$\angle ACB=90^\circ.$$

This is called the angle in a semicircle theorem. We can also say that the angle at the circumference subtended by a diameter is a right angle: the lines $CA$ and $CB$ from $C$ to the endpoints of the diameter form the subtended angle.

The position of $C$ can change around either semicircle, so the triangle can look tall, shallow or uneven. Provided that $AB$ remains a diameter and $C$ remains on the circumference, the angle at $C$ remains $90^\circ$.

Recognising when the theorem applies

Two conditions must both be visible or stated:

1. the line segment forming one side of the triangle is a diameter, so it passes through the centre and has endpoints on the circle;
2. the opposite vertex lies on the circumference.

A chord whose endpoints lie on the circle is not necessarily a diameter. If it does not pass through the centre, this particular theorem cannot be used.

Why the angle must be $90^\circ$

Let $O$ be the centre of the circle and draw the extra radius $OC$. Because $OA$, $OB$ and $OC$ are radii of the same circle, they have equal lengths.

In triangle $AOC$, $OA=OC$, so it is isosceles. Write its equal base angles as

$$\angle OAC=\angle ACO=\alpha.$$

Similarly, triangle $BOC$ is isosceles because $OB=OC$. Write

$$\angle OBC=\angle BCO=\beta.$$

Since $A$, $O$ and $B$ lie on one straight line, the angle at $A$ in triangle $ABC$ is $\alpha$, the angle at $B$ is $\beta$, and the complete angle at $C$ is $\alpha+\beta$. Using the angle sum of a triangle gives

$$\alpha+\beta+(\alpha+\beta)=180^\circ,$$

so

$$2(\alpha+\beta)=180^\circ \quad\Rightarrow\quad \alpha+\beta=90^\circ.$$

But $\angle ACB=\alpha+\beta$. Therefore $\angle ACB=90^\circ$, which proves the theorem.

The result also agrees with the centre–circumference angle theorem: a diameter makes a straight angle of $180^\circ$ at the centre, and the angle at the circumference subtended by the same endpoints is half of this, namely $90^\circ$.

Applying the theorem to find an angle

Suppose $ST$ is a diameter, $R$ lies on the circumference and $\angle RTS=34^\circ$. The diameter subtends $\angle SRT$, so

$$\angle SRT=90^\circ.$$

The angles in triangle $SRT$ total $180^\circ$, giving

$$\angle RST=180^\circ-90^\circ-34^\circ=56^\circ.$$

The theorem can also unlock length calculations. Since the triangle formed on a diameter is right-angled, Pythagoras’ theorem can be used when enough side lengths are known. For example, if the diameter is $5$ cm and one shorter side is $3$ cm, the other shorter side $x$ satisfies

$$x^2+3^2=5^2,$$

so $x=4$ cm. The diameter is the hypotenuse because it lies opposite the right angle.

Recall the theorem

If $AB$ is a diameter and $C$ is on the circumference, then

$$\angle ACB=90^\circ.$$

The right angle is opposite the diameter.

Conditions to check

• The relevant chord passes through the centre, so it is a diameter.
• The angle's vertex is on the circumference.

Proof skeleton

Draw $OC$. Since $OA=OB=OC$ (radii), triangles $AOC$ and $BOC$ are isosceles. Label their base angles $\alpha,\alpha$ and $\beta,\beta$. In triangle $ABC$:

$$\alpha+\beta+(\alpha+\beta)=180^\circ,$$

hence $\alpha+\beta=90^\circ$, so the angle subtended by the diameter at $C$ is a right angle.

Apply

Mark the $90^\circ$ angle first. Then use the triangle angle sum, algebra, trigonometry or Pythagoras as required. In the resulting right-angled triangle, the diameter is the hypotenuse.