Perpendicular Bisectors and the Circumcenter: Practice

Half of AB is 10 cm. Triangle PMA is right at M.

The theorem makes the endpoint distances equal.

Both lengths then equal 29, so they are positive.

All vertex distances from the circumcenter are equal.

The hypotenuse is twice the circumradius, so it is 10 cm. Use the known leg and hypotenuse in Pythagoras.

Let M be the midpoint of AB. Both compass crossings lie on its perpendicular bisector, on opposite sides of AB. Half the segment is 7 cm, and AX is a 25 cm radius. Pythagoras in right triangle AMX gives

The same calculation in triangle AMY gives a midpoint-to-crossing distance of 24 cm on the other side of AB. Add the two parts.

Both X and Y are equally far from A and B because the circles have equal radii. Both therefore lie on the perpendicular bisector, and their joining line is that line.

A circle centered at P gives equal distances to its two intersections with the line. Since those intersections lie on opposite sides, P is their midpoint.

With only T there is no nonzero segment AB whose perpendicular bisector can be used. Increasing the radius produces two distinct line intersections.

All three angles are acute. Applying the midpoint-parallel location argument to each side puts the center in all three interior half-planes.

Passing through the midpoint gives segment bisection. The 50 degree angle fails perpendicularity, which requires a right angle.

Each chord bisector contains the given rotation center by equidistance. Nonparallel chords produce nonparallel bisectors, so they have one intersection.

The concurrency proof makes all three center-to-vertex distances equal. Using any one of them as the radius draws the circle through every vertex.