Three marks are given on a flat plan. We need a point equally distant from all three. A “visual middle” does not guarantee it. Instead, turn the distance requirements into lines whose intersection provides an exact answer.
A locus consists of all points in a specified region satisfying a condition. Perpendicular bisectors locate the centre of a circle through a triangle’s vertices; internal angle bisectors locate the centre of a circle tangent to its sides. We need the three special segments and the definition of a circle. All points here lie in one Euclidean plane.
Check both directions
To call a line a locus, prove two things: every point on it satisfies the condition, and every point satisfying the condition lies on it. Otherwise we may lose solutions or add unwanted ones.
For example, all points at distance R>0 from O form a circle. The disk’s interior is excluded: its points are closer than R.
Equal distances to two marks
Let A and B be distinct, and M the midpoint of AB. The line through M perpendicular to AB is its perpendicular bisector, the locus of points P for which:
PA = PB.
For P on the line with P≠M, right triangles PMA and PMB have equal legs: PM is shared and MA=MB. Hence PA=PB. For M itself equality follows from the midpoint definition.
Conversely, if PA=PB and P is off AB, triangle PAB is isosceles. Its median PM is an altitude, so P is on the required perpendicular. If P lies on AB, equality occurs only at M. No points have been lost or added. AMSI construction and proof.
PA = PB: perpendicular bisector
PH = PK: internal angle bisector
O and I are different centres
Equal distances to an angle’s sides
For points inside a nonzero angle smaller than 180°, equal distances to its side lines define the internal angle bisector. Distance means a perpendicular length, not an arbitrary oblique segment.
In the diagram OP bisects the angle and PH,PK are equal perpendiculars. Equal acute angles and the shared hypotenuse OP make the right triangles congruent. Conversely, equal perpendicular legs and the common hypotenuse give equal angles at O. Restricting the region matters: elsewhere in the plane other branches of the angle-bisector lines also occur. Euclid IV.4, commentary.
A circle through three marks: the circumcircle
Draw the perpendicular bisectors of AB and BC in triangle ABC. They cannot be parallel: otherwise AB and BC, perpendicular to parallel lines, would themselves be parallel and would not form a triangle.
Let O be their unique intersection. Then OA=OB and OB=OC, so:
OA = OB = OC = R.
The circle of centre O and radius R passes through all vertices: it is the circumcircle. Another centre is impossible because it would lie at the same intersection. Euclid IV.5.
O lies inside an acute triangle, at the hypotenuse midpoint in a right triangle, and outside an obtuse triangle. No circle passes through three distinct collinear points: a line shares at most two points with a circle.
A circle between three sides: the incircle
Two internal bisectors meet inside the triangle at I. The half-angles at A and B are positive and sum to less than 90°, so their rays are not parallel. Their intersection belongs to both interior angles, hence lies in the triangle.
The angle-bisector property makes I equally distant from AB and AC, and from AB and BC: thus from all three side lines.
Call the shared perpendicular distance r. The circle with centre I and radius r lies inside and touches every side: it is the incircle. Equal distances to AC and BC also put I on the third internal bisector. Any other interior centre would be the same bisector intersection, so the circle is unique. Euclid IV.4.
I is always inside and generally differs from O. Equal distances to three vertices and equal distances to three sides are different tasks. Three marks call for O; a round boundary between three edges calls for I.