The angle bisector is the set of points equidistant from two lines
The last lesson of Phase 4 — the same locus idea as the one before it, aimed at lines instead of points.
The last lesson found the set of points equidistant from two points. This one asks the same kind of question about two lines instead: what is the set of points equidistant from both sides of an angle?
A fixed angle DOE at O, with its bisector drawn. A free point, P.
A fixed angle at O, with its bisector drawn, and a free point P. The distance from P to each of the two sides of the angle is unequal almost everywhere P is dragged, and equal exactly when, and only when, P sits on the drawn bisector. This needs JavaScript switched on.
This is the same snap-to-a-line behaviour the perpendicular bisector showed, except the anchors here are two lines instead of two points. A line that splits an angle into two equal angles is that angle’s bisector — already used, by name, to build a congruent pair of triangles two lessons ago. “Distance from a point to a line” means the length of the shortest path from the point to the line, which is always the perpendicular one. This lesson’s claim: the locus of points equidistant from both sides of an angle is exactly that angle’s bisector.
From the bisector to equal distances
Take a point P on the bisector, and drop a perpendicular to each side of the angle, landing at F on one side and G on the other. Triangle OPF and triangle OPG share the hypotenuse OP, have equal angles at O (the bisector splits the angle equally, by definition), and both carry a right angle, at F and at G. Two angles and a shared side: ASA, so the triangles are congruent, so PF equals PG — P is equidistant from both lines.
From equal distances back to the bisector
This reverse direction needs its own argument again (the same two-direction requirement as the last lesson). If P is equidistant from both lines, drop the same two perpendiculars, to F and to G. Triangle OPF and triangle OPG share hypotenuse OP, have equal legs PF and PG (given), and both carry a right angle at F and at G. A shared hypotenuse and one equal leg, with the right angle fixing where the second leg has to land, is enough to force the triangles congruent — a fourth criterion, alongside SSS, SAS and ASA, that a right angle earns for free. Congruence hands over the angles at O, equal on both sides, so OP bisects angle DOE.
Put side by side, “equidistant from two points” and “equidistant from two lines” turn out to be the same kind of question, answered the same way, even though a point and a line feel like different kinds of objects. Recognising when two problems share a shape, even while looking different on the surface, is worth more than solving either one alone — it is what lets a proof already built for one situation be adapted, rather than reinvented, for the next.
This closes Phase 4. Phase 5, quadrilaterals, reuses SSS, SAS, ASA and the isosceles pair immediately: a parallelogram’s opposite sides being equal is proved with a single diagonal and one pair of congruent triangles, the same move this whole phase has practised.
P is equidistant from the two sides of an angle. Is P necessarily on the segment between the two points where those perpendiculars land?
No — P only needs to sit on the bisector ray from the vertex. Where exactly it sits along that ray is free; only its distance to each side has to match.
Two angle bisectors of a triangle cross at a point. What can be said about that point’s distance to all three sides?
It is equidistant from all three — each bisector guarantees equal distance to the two sides it splits, and where two bisectors cross, both guarantees hold at once, forcing the distance to all three sides to agree.
Why does this lesson’s proof route through two right-angled triangles, OPF and OPG, rather than splitting the angle directly the way G4.3 split a triangle?
Because “distance to a line” is defined as a perpendicular length, so any proof about it has to build that perpendicular somewhere. Dropping it from P to each side is what turns “equal distance” into “equal side length in a triangle”, which is the only form SSS, SAS and ASA know how to use.
This lesson counts as done once you get its exercise right. Nothing to tick off by hand.