The exterior angle theorem
What happens to a triangle's angle sum when one side is extended past a vertex — and why the last lesson already proved it.
Extend one side of a triangle past a vertex, and a new angle appears outside the triangle, between the extension and the next side. This lesson is about what that angle is worth — and it turns out the last lesson already worked out the answer, without knowing it yet.
The same triangle as before, A, B and C, all draggable, with side AB extended past B.
A triangle, A, B and C, all three draggable, with side AB extended beyond B. The angle between that extension and side BC — the exterior angle at B — always equals the sum of the two interior angles it is not next to, the ones at A and at C, whatever shape the triangle is dragged into. This needs JavaScript switched on.
That match is not a coincidence of measurement — the last lesson already proves it, as the reasoning below shows.
Naming the new angle
The angle between a side extended past a vertex and the triangle’s next side is called an exterior angle. The two interior angles that are not at that same vertex — here, the angles at A and at C — are its remote interior angles: “remote” because neither sits next to where the exterior angle was formed.
Reading its value off the last lesson
Extending AB past B does not change the direction of AB: the extension runs the same way, just further. So the exterior angle at B and the interior angle at B sit on either side of one straight line — the original side AB, continued — and a straight line’s angles sum to 180 (the very first lesson of this course), so the exterior angle is 180 minus the interior angle at B.
The last lesson proved the interior angle at B, plus the interior angle at A, plus the interior angle at C, sum to 180. Rearranged, the interior angle at B is 180 minus the other two. Substitute that into “exterior angle at B is 180 minus the interior angle at B”, and the 180s cancel: the exterior angle at B equals the interior angle at A plus the interior angle at C. Nothing new was assumed. Both steps were already proved.
Recognising when a new-looking claim is actually an old one, rearranged, is most of what doing geometry well consists of. The exterior angle theorem is often stated as if it were its own idea; here it is visibly a repackaging of the last two lessons, which is the more useful thing to notice.
Watching the exterior angle theorem be built entirely from the last lesson, without a single new measurement, is the same habit the angle sum of any polygon needs next: that theorem, too, is one new idea (splitting a polygon into triangles) applied to an old one, rather than a new proof from nothing.
A triangle’s exterior angle at one vertex is 110 degrees. What can you say about the two remote interior angles?
They sum to 110 degrees, whatever they are individually — the theorem fixes their sum, not their split. Many different triangles share that same exterior angle.
Why must the exterior angle always be bigger than either remote interior angle on its own?
Because it equals their sum, and an interior angle of a triangle is always a positive angle greater than zero degrees. Adding a positive amount to one of them can only make the total bigger than that one alone.
Could this theorem be proved a different way, without going through the triangle angle sum first?
Yes — directly, by drawing a line through B parallel to AC and reading alternate and corresponding angles off it. That is a second, independent route to the same fact. This lesson took the shorter path, through a theorem already sitting on the shelf, because it was already there to use.
This lesson counts as done once you get its exercise right. Nothing to tick off by hand.