Angle in a semicircle is 90 degrees

We can draw a triangle ABC where AB is a diameter of a circle, and C lies on the circumference of the circle:

The angle at the circumference C is a right angle. We say the angle in a semicircle is a right angle.

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Proof

We can prove this as follows.

We start by drawing an extra line from the centre O to the point C.

Notice that the lines OA, OB and OC are all radii of the circle, and therefore all of equal length.

Looking at the triangle AOC, this is an isosceles triangle (from the rule 2 radii form an isosceles triangle). So the two angles at the circumference are equal (we will call them a):

By the same logic, the two angles at the circumference in triangle BOC are equal, and we will call them b:

Looking at the original triangle ABC:

The angle at A is a, the angle at B is b, and the angle at C is a+b. Since the three angles of a triangle add up to 180° we have:

Gathering the terms a and b:

Dividing both sides by 2 gives:

Since the angle at C is a + b, this proves that the angle at C is 90°.

See also

• Parts of a circle
• Tangent and radius of a circle meet at 90°
• Two radii form an isosceles triangle
• Perpendicular bisector of a chord
• Angle at the centre of a circle is twice the angle at the circumference
• Angles in the same segment of a circle are equal
• Opposite angles in a cyclic quadrilateral add up to 180°
• Two tangents from a point have equal length