The triangle inequality
The last lesson of Phase 2 — and the only one that needed nothing from inside it at all.
Every lesson so far in this phase has built on the one before it. This one does not. It reaches all the way back to the very first lesson of the course — point, line, and what “straight” means — and needs nothing else.
A triangle, A, B and C, all draggable.
A triangle, A, B and C, all three draggable. Going by way of B — A to B, then B to C — is never shorter than going straight from A to C, whatever shape the triangle is dragged into, and the two match exactly only when B sits precisely on the straight line between A and C. This needs JavaScript switched on.
Sometimes the detour is a lot longer; sometimes, with B close to the line between A and C, only a little. Try to drag B so the detour matches the direct route exactly — it only happens in one place.
A triangle where one vertex sits exactly on the line through the other two is called degenerate: not really a triangle any more, since it has folded flat, with no area and no angles worth measuring. It sits at the exact boundary of the triangle inequality — the one configuration where the detour and the direct distance are equal instead of the detour being longer.
The one fact this needs
This lesson needs no auxiliary line, no diagonal, no earlier angle theorem. It needs one fact from the very first lesson of the course: a straight line is the shortest path between two points. The distance from A to C, measured directly, is by definition the shortest any path between them can be. A path that goes from A to B and then to C is one particular path between A and C — just not usually the straight one — so it can never beat the direct distance, only match or exceed it.
It matches only when the detour secretly is the direct route: when B lies on the segment from A to C, going A to B to C traces out the same straight line twice, in two pieces, adding up to exactly the same length as going straight through. Move B off that line in any direction, and the detour stops being straight, and the shortest-path fact guarantees it can only get longer, never shorter.
Not every theorem needs the machinery this phase has been building: sometimes the shortest proof was available on lesson one, and the useful skill is recognising that, rather than reaching for the newest tool by habit.
Phase 3, congruence, asks a different kind of question: not “how long can a path be”, but “when do two triangles have to be identical”. The triangle inequality reappears there as a constraint on which three lengths can even form a triangle in the first place, before congruence asks whether two particular ones are the same shape.
Can three lengths of 3, 4 and 9 units form a triangle?
No. The two shorter sides, 3 and 4, sum to 7, which is less than the third side, 9 — there is no way to swing a 3-unit side and a 4-unit side around so their far ends meet 9 units apart. The detour would have to be shorter than the direct distance, which this lesson just showed is impossible.
Can three lengths of 3, 4 and 7 units form a triangle?
Only the degenerate one: 3 plus 4 equals exactly 7, so the three points must lie on one straight line, with no area at all. It sits exactly on the boundary this lesson found, not inside it.
Why does a phase about triangles end on a lesson that barely mentions angles?
Because a triangle is not only a shape with three angles — it is also a shape with three lengths, and a claim can be true about one without needing the other. Ending here is a reminder that this course’s tools (parallel lines, angle sums) are not the only tools geometry has; distance and straightness were enough on their own, for this one.
This lesson counts as done once you get its exercise right. Nothing to tick off by hand.