12 multiple-choice questions, progressively harder.
The endpoints A and B coincide. What is the locus of points P with equal distances to A and B?
Solution
Correct answer: D
When A and B are the same point, the two measured distances are identical for every P. Distinct endpoints are essential to the perpendicular-bisector locus theorem.
In the converse proof, P is off AB and has equal distances to A and B. M is the midpoint. Which congruence test proves triangles PMA and PMB congruent before any right angle is known?
Correct answer: B
The endpoint distances match, the two halves at M match, and PM is common. SSS therefore applies without assuming the perpendicularity being proved.
A point P on the perpendicular bisector is the midpoint M itself. What is the valid proof of equal endpoint distances?
Correct answer: A
At P equal to M, PM has zero length, so the proposed triangles are degenerate. The midpoint definition already gives equal endpoint distances.
Why do the perpendicular bisectors of AB and AC intersect when ABC is a nondegenerate triangle?
Correct answer: C
The sides AB and AC are distinct intersecting lines, so they are not parallel. Parallel perpendicular bisectors would force those sides to be parallel, a contradiction.
Two arbitrary point-image pairs each have equal endpoint distances from O. About O, the first pair requires a 40 degree counterclockwise turn; the second requires a 70 degree counterclockwise turn. What follows?
A rotation about O uses one common directed angle. These two mappings require different counterclockwise turns about O, 40 degrees and 70 degrees, so one rotation about O cannot perform both.
Two pairs come from one given nonidentity rotation. Their nonzero point-image chords are parallel. Why can these pairs fail to locate one center by intersecting distinct bisectors?
Perpendiculars to parallel chords are parallel. Since both bisectors contain the already-given center, they cannot be distinct parallel lines; they are the same line. Choose another suitable pair.
A triangle has a 104 degree angle at C. Which claim about its circumcenter O is justified?
The angle at C is obtuse, so AB is the longest side. In the midpoint-parallel proof, the angle at the AC midpoint exceeds 90 degrees and places O across AB from C.
In an acute triangle, why is it insufficient to prove only that O is on the same side of AB as C when claiming O is inside?
One half-plane contains many points outside the triangle. Repeating the acute-angle argument for all three sides puts O in their intersection, which is the interior.
Three distinct points A, B, and C lie in that order on a line. Can a circle pass through all three?
Any such center would have to lie on both bisectors by the converse. Those lines are perpendicular to the same line at different midpoints, so they are distinct and parallel. They have no common point.
P is off line AB and has equal distances to A and B. Circles centered at A and B use radius AP. Why is there a second intersection besides P?
Since PA and PB have the same length, the strict triangle inequality gives
Thus the equal radius AP exceeds half AB, giving two distinct crossings. P is one, and the other lies across AB.
A proposed circumcircle has radius 5 cm for a triangle with a 12 cm side AB. Which conclusion is justified?
If O were the center, OA and OB would each be 5 cm. The triangle inequality, including the straight-line boundary case, requires
A 12 cm side violates this bound, so that proposed circle cannot pass through both endpoints.
Three noncollinear points are given with no circle drawn. What establishes that their circumcircle is unique?
The converse forces every possible center onto the same two intersecting bisectors. Their one intersection fixes the center, and its distance to a vertex fixes the radius.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.