Why Every Angle Drawn in a Semicircle Is Exactly a Right Angle

Take two points $A$ and $B$ on a circle. The part of the circle between them is an arc, and the straight line $AB$ is a chord. The chord divides the circle into two segments.

Join $A$ and $B$ to another point — the centre $O$, or a point $C$ on the circle — and the angle formed is the angle subtended by the arc at that point.

This part covers four circle angle results and mixed problems.

**The angle an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circle: $\angle AOB=2\angle ACB$.

Why it works.** Join $C$ to $O$ and extend it. Since $OA=OC=OB$, triangles $OAC$ and $OBC$ are isosceles, and each exterior angle at $O$ equals twice a base angle. Adding the two parts gives $\angle AOB=2\angle ACB$.

Worked example 1. $\angle ACB=35^{\circ }$, so $\angle AOB=70^{\circ }$.

Worked example 2. $\angle AOB=130^{\circ }$, so $\angle ACB=65^{\circ }$.

Worked example 3 — reflex angle. The minor $\angle AOB=100^{\circ }$, and $D$ lies on the minor arc $AB$. The arc on the other side subtends the reflex angle $360^{\circ }-100^{\circ }=260^{\circ }$ at $O$, so

$$\angle ADB=\frac{260^{\circ }}{2}=130^{\circ }$$

Worked example 4. With $\angle AOB=70^{\circ }$ and $OA=OB$:

$$\angle OAB=\frac{180^{\circ }-70^{\circ }}{2}=55^{\circ }$$

An everyday example. In a circular park with a statue at the centre, two gates on the boundary appear at twice the angle from the statue that they do from a bench on the far edge.

The substance. Always check which arc the angle stands on — the reflex angle at the centre goes with points on the minor arc.

Angles subtended by the same arc at points in the same segment of a circle are equal, because each is half of the same angle at the centre.

Worked example 1. $C$ and $D$ lie on the same side of chord $AB$, and $\angle ACB=40^{\circ }$. Then $\angle ADB=40^{\circ }$.

Worked example 2. Chords $AC$ and $BD$ of a circle cross at $E$. $\angle BAC=30^{\circ }$ and $\angle ABD=50^{\circ }$.

- $\angle BDC=\angle BAC=30^{\circ }$ — both stand on arc $BC$
- $\angle ACD=\angle ABD=50^{\circ }$ — both stand on arc $AD$

Worked example 3. In the same figure, the angles of $△ABE$ give

$$\angle AEB=180^{\circ }-30^{\circ }-50^{\circ }=100^{\circ }$$

An everyday example. People anywhere along one curved edge of a circular lake see a straight footbridge across it at the same angle.

The boundary case. **Points in opposite segments give angles that add to $180^{\circ }$**, not equal angles — the idea behind cyclic quadrilaterals in Part 2.

**If $AB$ is a diameter and $C$ is any other point on the circle, then $\angle ACB=90^{\circ }$, because the angle at the centre is a straight angle of $180^{\circ }$.

Worked example 1 — angles.** $AB$ is a diameter and $\angle CAB=35^{\circ }$.

$$\angle ACB=90^{\circ }\angle ABC=180^{\circ }-90^{\circ }-35^{\circ }=55^{\circ }$$

Worked example 2 — lengths. $AB=13$ cm is a diameter and $AC=5$ cm.

$$BC=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\text{ cm}$$

Worked example 3 — radius. A right triangle with legs $6$ cm and $8$ cm is drawn inside a circle with all vertices on it.

$$\text{hypotenuse}=\sqrt{36+64}=10\text{ cm}=\text{diameter}\text{radius}=5\text{ cm}$$

An everyday example. To find the centre of a round wooden tabletop, place the right-angled corner of a set square on its edge and mark where the two arms cut the rim; that line is a diameter. Doing this twice, the diameters cross at the centre.

The substance. The converse also holds: if a triangle's vertices lie on a circle and one angle is $90^{\circ }$, the opposite side is a diameter.

Mark equal radii to find isosceles triangles, apply the circle theorems to link angles at the centre and circumference, and finish with angle sums of triangles or around a point.

Worked example 1. $O$ is the centre and $\angle OAB=25^{\circ }$.

$$\angle OBA=25^{\circ }(OA=OB)\angle AOB=180^{\circ }-50^{\circ }=130^{\circ }\angle ACB=65^{\circ }$$

Worked example 2. $A$, $B$, $C$ lie on a circle with centre $O$, $\angle AOB=140^{\circ }$ and $\angle BOC=100^{\circ }$.

$$\angle AOC=360^{\circ }-140^{\circ }-100^{\circ }=120^{\circ }$$
$$\angle ABC=60^{\circ }\angle BAC=50^{\circ }\angle BCA=70^{\circ }$$

Check: $60^{\circ }+50^{\circ }+70^{\circ }=180^{\circ }$.

Worked example 3. $AB$ is a diameter, $\angle BAC=28^{\circ }$, and $D$ lies on the same arc as $B$ with respect to chord $AC$.

$$\angle ABC=90^{\circ }-28^{\circ }=62^{\circ }\angle ADC=\angle ABC=62^{\circ }$$

An everyday example. A circular rangoli with a centre dot and spokes creates exactly these isosceles triangles.

The substance. Radii are the hidden equal sides in almost every circle angle problem.