Circle theorems: chords, tangents and inscribed angles

Archaeologists have found a small piece of an ancient plate. How can they tell the size of the whole plate? It is enough to draw two chords along the edge of the piece and construct their perpendicular bisectors: they meet at the centre of the plate. In this lesson we learn such “hidden” properties of the circle: chords and tangents, central and inscribed angles, and the circles drawn around and inside a triangle.

In the lesson “Circumference and area of a circle” we met the circle, its radius and diameter and the length of an arc. Here our main tools are congruent triangles (“Congruent triangles”) and similar triangles (“Similar triangles”).

Chords and diameters

Join the centre O to the ends A and B of a chord: since OA = OB = R, triangle AOB is isosceles, and its altitude OH is also a median. This gives three properties: a diameter perpendicular to a chord bisects it; the perpendicular bisector of a chord passes through the centre; equal chords are equally far from the centre. The right triangle OAH gives a formula linking the length of a chord to its distance from the centre.

Finding the centre of the broken plate — the perpendicular bisector of every chord passes through the centre:

1. 1
   Two chords

   Take three points A, B, C on the arc and draw chords AB and BC.

2. 2
   Perpendicular bisectors

   Construct the perpendicular bisectors of AB and BC with compass and ruler (as in the lesson “Congruent triangles”).

3. 3
   The centre

   The bisectors meet at the centre O, and OA is the radius.

Tangents and their properties

Let d be the distance from the centre to a line. If d < R, the line crosses the circle at two points — it is a secant; if d > R, they have no common points; if d = R, there is exactly one common point — the line touches the circle. A tangent is perpendicular to the radius drawn to the point of tangency: if radius OA were not perpendicular, the perpendicular from O to the line would be shorter than OA, that is, shorter than R, and the line would cross the circle at two points. From a point P outside the circle you can draw two tangents. The right triangles PAO and PBO have OA = OB = R and the common hypotenuse PO, so they are congruent (hypotenuse and leg).

Central and inscribed angles

Proof (the simple case: side CB of the angle is a diameter, so the centre O lies on it). OA = OC, so triangle AOC is isosceles and ∠OAC = ∠OCA. Angle AOB is an exterior angle of this triangle and equals the sum of the two interior angles not adjacent to it: ∠AOB = 2∠ACB. In general, draw the diameter CD from C: if O lies inside the angle, the diameter splits it into two parts, the simple case holds for each part, and we add the equalities; if O lies outside the angle, we subtract them. Two important consequences: inscribed angles that subtend the same arc are equal; an inscribed angle that subtends a diameter is a right angle (a semicircle is 180°, half of it is 90°).

Segments of chords, secants and tangents

Let chords AB and CD meet at P. In △APC and △DPB, ∠APC = ∠DPB (vertical angles), and ∠CAB = ∠CDB are inscribed angles on the same arc CB. By AA, △APC ∼ △DPB, so AP / DP = CP / BP, that is, AP · PB = CP · PD. The same argument works for secants drawn from a point P outside the circle; when one secant turns into a tangent, its two intersection points merge and the product becomes the square of the tangent.

Inscribed and circumscribed circles

The circle through the vertices of a triangle is its circumscribed circle (circumcircle). Its centre is equally far from every vertex, so it is the point where the perpendicular bisectors of the sides meet (all three meet at one point). The circle that touches all three sides is the inscribed circle (incircle); its centre is equally far from the sides, so it is the point where the angle bisectors meet. Why do the three perpendicular bisectors meet at one point? Let the perpendicular bisectors of AB and BC meet at O: then OA = OB and OB = OC, so OA = OC and O lies on the third perpendicular bisector too. The same argument with distances to the sides works for the angle bisectors. In an acute triangle O lies inside the triangle, in a right triangle at the midpoint of the hypotenuse, and in an obtuse triangle outside it.

If all the vertices of a quadrilateral lie on a circle, it is a cyclic quadrilateral. Its opposite angles (for example A and C) together subtend the whole circle, so ∠A + ∠C = ½ · 360° = 180°. A quadrilateral whose sides all touch a circle is a tangential quadrilateral; since the two tangent segments from each vertex are equal, the sums of opposite sides are equal. The converse of each property is true too (for convex quadrilaterals).