12 multiple-choice questions, progressively harder.
Why is the altitude-concurrence theorem stated for a nondegenerate triangle?
Solution
Correct answer: A
Collinear vertices do not form the required triangular region or three distinct side directions. The auxiliary-triangle and perpendicular-bisector proof requires a genuine triangle.
Must the three vertex-to-foot altitude segments themselves all contain the orthocenter?
Correct answer: D
Concurrence is a theorem about the complete altitude lines. The orthocenter of an obtuse triangle lies outside the triangle and beyond some drawn vertex-to-foot segments.
Which triangle receives the perpendicular-bisector concurrence theorem in the altitude proof?
Correct answer: B
The original vertices become midpoints of the auxiliary sides. The original altitude lines are perpendicular bisectors of those auxiliary sides, so the theorem applies to the larger triangle.
Which pair of centers must lie strictly inside every nondegenerate triangle?
Correct answer: C
The internal-bisector and median theorems put the incenter and centroid inside. Both the orthocenter and circumcenter can be outside in an obtuse triangle.
Why are the orthocenter and circumcenter distinct in a nondegenerate right triangle?
The right-angle vertex is not on its opposite hypotenuse. The hypotenuse midpoint and that vertex are therefore distinct points.
Why is the auxiliary side QR through A not itself the altitude line from A?
QR has the same direction as BC. The altitude from A has a perpendicular direction. One line cannot have both directions at once.
In the auxiliary construction, which parallelogram directly gives RB the same length as AC?
In quadrilateral ACBR, AC and BR are opposite sides. Its two parallel pairs establish a parallelogram, so
Can two altitude feet coincide in a nondegenerate triangle?
In a triangle right at C, the altitudes from A and B are AC and BC. Both feet are C. The third foot lies on AB.
An obtuse triangle has its obtuse angle at A. How many altitude feet lie on side extensions rather than inside the side segments?
The feet from B and C lie beyond A on extensions. The foot from A lies inside BC because the other two triangle angles are acute.
In an obtuse triangle, can its orthocenter equal its incenter or its centroid?
The orthocenter is outside an obtuse triangle. The incenter and centroid are both strictly inside, so neither can be the same point.
For a given equilateral triangle, which procedure locates the common point of all four centers?
In an equilateral triangle the altitude, median, angle-bisector and perpendicular-bisector line families coincide. Two altitude lines therefore locate their shared center.
ABC is right at C. In its auxiliary triangle PQR, A lies on QR, B on RP, and C on PQ. Which auxiliary side has midpoint H, the original orthocenter?
The original orthocenter is C. Since the construction makes C the midpoint of PQ, H is the midpoint of PQ. Also QR is parallel to BC and RP to CA, so the auxiliary angle at R is right and PQ is its hypotenuse.
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