If two tangents are drawn from a point T to a circle with centre O, and P and R are the points of contact of the tangents with the circle, then, using symmetry,

(a)PT = RT

(b)Triangles TPO and TRO are congruent.

The angle between a tangent and a chord equals an angle at the circumference subtended by the same chord;

e.g. a = b in the diagram.

This is known as the

alternate segment theorem

and needs a proof, as it is not obviously true!

Construct the diameter POS, as shown.

We know that

angle SRP = 90°

since PS is a diameter.

Now

angle PSR = angle PQR = x°, say,

so

angle SPR	= 180° − 90° − x
	= 90° − x°

But

angle RPT	= 90° − (angle SPR)
	= 90° − (90° − x°)
	= x°
	= angle PQR

and the result is proved.

For any two intersecting chords, as shown,

The proof is based on similar triangles.

In triangles AXB and DXC,

angle BAC = angle BDC (equal angles subtended by chord BC)

and

angle ABD = angle ACD (equal angles subtended by chord AD)

As AXB and DXC are similar,

=   ⇒   AX.CX = BX.DX

as required.

This result will still be true even when the chords intersect outside the circle, as illustrated opposite.

How can this be proved?

When the chord BD becomes a tangent, and B and D coincide at the point P, then

AX × CX = PX × PX

or