Angle Properties of Circles - Theorems 3 and 4
Theory
The angle in a semicircle is a right angle (Theorem 3), and the opposite angles of a cyclic quadrilateral add to \(180^\circ\) (Theorem 4). This Year 10A maths page shows how to use them to find angles and lengths, with algebra, Pythagoras' theorem and short proofs, giving a reason at every step.
A diameter cuts a circle into two semicircles. An angle in a semicircle has its vertex on the circle and its arms passing through the two ends of a diameter.
A cyclic quadrilateral has all four vertices on a circle. Its opposite angles are the pairs that do not share a side: \(\angle A\) with \(\angle C\), and \(\angle B\) with \(\angle D\). Its exterior angle is made by producing (extending) one side.
Theorem 3: the angle in a semicircle is a right angle. Theorem 4: the opposite angles of a cyclic quadrilateral are supplementary (add to \(180^\circ\)). Both come from Theorem 1: a diameter makes a straight angle of \(180^\circ\) at the centre, and two opposite angles of a cyclic quadrilateral stand on arcs that together make a full revolution at the centre.
Every answer needs a reason, written as a short theorem name such as "angle in a semicircle" or "opposite angles of a cyclic quadrilateral".
It prepares you for senior maths courses, where you prove results as well as use them.
If \(AB\) is a diameter and \(C\) is any other point on the circle:
For a cyclic quadrilateral \(ABCD\):
| Theorem | Reason to write |
|---|---|
| 3. Angle in a semicircle \(= 90^\circ\) | angle in a semicircle |
| 4. Opposite angles add to \(180^\circ\) | opposite angles of a cyclic quadrilateral |
| Exterior angle \(=\) interior opposite angle | exterior angle of a cyclic quadrilateral |
How to find an angle or a length
- Mark the given angles, any diameter and any equal radii on the diagram.
- Look for a diameter: the angle opposite it, on the circle, is \(90^\circ\). Look for four points on the circle joined in order: opposite angles add to \(180^\circ\).
- Calculate one angle at a time, with the reason beside each step.
Problem types
- Expressions in a pronumeral: opposite angles add to \(180\), or the exterior angle equals the interior opposite angle. Solve, then substitute back (Example 1).
- Two unknowns: each pair of opposite angles gives one equation. Solve the two equations simultaneously, then check that each pair adds to \(180^\circ\).
- Ratios and differences: if \(\angle B : \angle D = 4 : 5\), write \(4k\) and \(5k\) with \(4k + 5k = 180\). If one angle is \(30^\circ\) more than its opposite angle, write \(a\) and \(a + 30\) (Example 3).
- Lengths with a diameter: the triangle on the diameter is right-angled, so use Pythagoras' theorem. If the sides are expressions, expand, collect into a quadratic, factorise and reject the negative solution (Example 2).
- Angles with a diameter: in \(\triangle ABC\) with diameter \(AB\), the two angles at \(A\) and \(B\) add to \(90^\circ\).
- Proofs: name the angles, use Theorem 3 or 4 with the properties of the shape (parallel sides give co-interior angles; a parallelogram has equal opposite angles), and finish with the statement to be proved (Example 4). To prove Theorem 4 itself, join \(O\) to two opposite vertices: the angles at \(O\) are \(2x\) and \(2y\) by Theorem 1, and they make a revolution. To prove Theorem 3, join \(O\) to the point on the circle and use two isosceles triangles.
Opposite angles of a cyclic quadrilateral are supplementary:
| \((5p - 20) + (2p + 18)\) | \(=\) | \(180\) |
| \(7p - 2\) | \(=\) | \(180\) |
| \(p\) | \(=\) | \(\mathbf{26}\) |
So \(\angle DAB = 5(26) - 20 = 110^\circ\). The exterior angle of a cyclic quadrilateral equals the interior opposite angle:
| \(\angle DCE\) | \(=\) | \(\angle DAB = \mathbf{110^\circ}\) |
Check: \(\angle BCD = 2(26) + 18 = 70^\circ\), and \(70^\circ + 110^\circ = 180^\circ\) on the straight line \(BCE\).
\(\angle ACB = 90^\circ\) (angle in a semicircle), so Pythagoras' theorem applies in \(\triangle ABC\):
| \(x^2 + (x + 14)^2\) | \(=\) | \(26^2\) |
| \(2x^2 + 28x + 196\) | \(=\) | \(676\) |
| \(x^2 + 14x - 240\) | \(=\) | \(0\) |
| \((x + 24)(x - 10)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(10 \ \text{or} \ -24\) |
A length cannot be negative, so \(x = \mathbf{10}\) and \(BC = 10 + 14 = \mathbf{24}\) cm.
Check: \(10^2 + 24^2 = 100 + 576 = 676 = 26^2\) \(\checkmark\)
\(\angle B\) and \(\angle D\) are opposite angles of a cyclic quadrilateral, so they add to \(180^\circ\). Write them as \(4k\) and \(5k\):
| \(4k + 5k\) | \(=\) | \(180^\circ\) |
| \(k\) | \(=\) | \(20^\circ\) |
| \(\angle B = 4k\) | \(=\) | \(\mathbf{80^\circ}\) |
| \(\angle D = 5k\) | \(=\) | \(\mathbf{100^\circ}\) |
\(\angle A\) and \(\angle C\) are also opposite angles. Let \(\angle A = a\), so \(\angle C = a + 30^\circ\):
| \(a + (a + 30^\circ)\) | \(=\) | \(180^\circ\) |
| \(2a\) | \(=\) | \(150^\circ\) |
| \(\angle A = a\) | \(=\) | \(\mathbf{75^\circ}\) |
| \(\angle C\) | \(=\) | \(\mathbf{105^\circ}\) |
Check: \(75^\circ + 80^\circ + 105^\circ + 100^\circ = 360^\circ\), the angle sum of a quadrilateral \(\checkmark\)
Opposite angles of a parallelogram are equal:
| \(\angle DAB\) | \(=\) | \(\angle BCD\) |
\(ABCD\) is a cyclic quadrilateral, so its opposite angles are supplementary:
| \(\angle DAB + \angle BCD\) | \(=\) | \(180^\circ\) |
| \(2\angle DAB\) | \(=\) | \(180^\circ\) |
| \(\angle DAB\) | \(=\) | \(90^\circ\) |
In the same way, \(\angle ABC = \angle CDA = 90^\circ\), and then \(\angle BCD = 90^\circ\) too. A parallelogram with four right angles is a rectangle. \(\blacksquare\)
Common pitfalls
Frequently asked questions
Why is the angle in a semicircle 90 degrees?
The diameter makes an angle of \(180^\circ\) at the centre, and the angle at the circumference standing on the same arc is half of that, \(90^\circ\).
What are the properties of a cyclic quadrilateral?
All four vertices lie on a circle. Opposite angles add to \(180^\circ\), and an exterior angle equals the interior opposite angle.
Is every quadrilateral cyclic?
No. A quadrilateral is cyclic only if its opposite angles add to \(180^\circ\). Every rectangle is cyclic, but a parallelogram that is not a rectangle is not.
How do you use Pythagoras' theorem with a diameter?
The angle opposite the diameter is a right angle, so the diameter is the hypotenuse. The squares of the other two sides add to the square of the diameter.
Do I need to give reasons in circle geometry?
Yes. Each step should name the theorem or fact used, such as angle in a semicircle, opposite angles of a cyclic quadrilateral, or angle sum of a triangle. Marks are given for the reasons as well as the answer.