The (Amazing) Alternate Segment Theorem!

In Savage Shapes there's a chapter on some of the strange things that happen with circles - but sadly there wasn't room to put everything. A Murderous Maths fan called ISLA KENNEDY asked us about "The Alternate Segment Theorem" so here it is...

Here's the funny bit:

The angle between the chord and the tangent is ALWAYS equal to the angle "in the alternate segment".

In other words, in our diagram, angle a is always the same as angle b.

If you're not sure what segments or chords are look at our guide to Names of Bits of Circles


If you've read Savage Shapes, you'll know that you're not allowed just to say something without explaining why it always works, so here's the proof. (You might want your copy of Savage Shapes handy to check on some of the details we're about to use.)

We've drawn a couple of radii (i.e. lines from the centre of the circle, we've made them red for clarity) to the ends of the chord. This makes an isosceles triangle with two equal angles which we've marked "x". (See page 104 of Savage Shapes.)

We split the isosceles triangle into two identical little right angled triangles. Each of them has an angle y at the centre of the circle.

The angle between the radius and the tangent is 90º (see page 114).

Now we just play with the angles a bit:

Go THIS way Go THAT way


You can go on to show that this theorem works for the angles over 90º too!

In this case, now we know that a = b we can show that s = t.

This is clever stuff - so if you've understood it WELL DONE!

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