12 multiple-choice questions, progressively harder.
In the HL overlay, why must C, A, and F be collinear after the third vertices are put on opposite sides of AB?
Solution
Correct answer: B
The right-angle givens place both third-vertex rays on the one perpendicular to AB at A. Choosing opposite half-planes makes them opposite rays, with A between C and F.
Which statement would make the isosceles-based HL proof circular?
Correct answer: D
Equality of the other legs is a conclusion of HL, not one of its givens. Assuming it would supply the missing equality the proof is supposed to establish.
Angles A and B in triangle ABC match angles D and E in triangle DEF. A record also says
Why does this record not by itself give AAS under the stated matching?
Correct answer: A
A corresponds to D, B to E, and C to F. Therefore AC maps to DF. An equality involving the different side EF does not supply the required corresponding side pair.
Two right triangles are right at Q and S. The given hypotenuses satisfy
The given legs satisfy
Which ordered congruence statement follows?
Correct answer: C
The right angles match Q to S. The matching legs PQ and RS then match P to R, leaving R of the first triangle matched with T of the second. HL gives
Triangles ABC and DEF have the following data:
Which pair of angle vertices can be used with the equal side in an ASA proof?
The third angles have the same measure:
AC is between angles A and C, and DF is between D and F. Those angle pairs provide the ASA route.
A construction yields two triangles with
and the same angle at A. Their bases AB1 and AB2 are unequal. What is the most direct reason these triangles are noncongruent under the shared-vertex correspondence?
The given equalities cover only two side pairs. The different base lengths violate a necessary condition for congruence.
The triangles share two side lengths, while their third lengths differ. Their full side-length lists therefore cannot match under a different vertex order either.
A right triangle has legs 12 and 16. Another right triangle has a leg of 12, but its hypotenuse length is not given. Is the one equal leg pair enough to prove congruence?
There is only one equal side pair and one equal angle pair. HL needs equal hypotenuses as well; SAS on the legs would need an equal second leg.
In the HL proof, once both small angles at B have been shown equal, which side pairs complete SAS?
The angle at B lies between AB and BC in one triangle, and between AB and BF in the other. Those side pairs were already known equal.
Which extra fact is sufficient to turn an AAA comparison into a congruence proof when all three angle pairs have a fixed correspondence?
Choose the matching angles at the endpoints of the added equal side pair. ASA now proves congruence.
For fixed SSA data, a construction circle touches the given ray at exactly one point beyond its endpoint. How many nondegenerate triangles does that intersection supply?
One intersection supplies one possible third vertex on the ray. SSA failing as a general guarantee does not mean every construction must have two solutions.
Could triangles satisfying unrestricted SSA measurements still be congruent?
Insufficient information leaves more than one possibility. It does not forbid a congruent pair from meeting the given equalities.
Two AAS-congruent triangles have corresponding angles (6u−7)∘(6u-7)^\circ(6u−7)∘ and (3u+20)∘(3u+20)^\circ(3u+20)∘. What is that common angle?
Corresponding angles in congruent triangles match.
Solving gives
Substitute into the numerical expression:
The common angle therefore has measure
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