Triangle Centers and Constructions: Chapter Review

Vocabulary and notation

Perpendicular bisector

The full line perpendicular to a segment through its midpoint. A segment bisector only passes through the midpoint; it need not be perpendicular.

Locus

The complete set of points satisfying a condition. Both theorem directions are needed to show that a proposed set has every required point and no extra points.

Supporting side line and perpendicular foot

The supporting line contains the entire side and both extensions. A perpendicular foot is where the perpendicular meets that line; it may miss the side segment.

Internal angle bisector

The ray dividing an angle into two equal interior angles. Its equal-distance property concerns perpendicular distances to the two supporting side lines.

Concurrent lines

Lines passing through one common point. Constructing two defining lines locates a candidate; the concurrence theorem places the third through that point.

Circumcenter $O$, circumcircle, circumradius $R$

The side perpendicular bisectors meet at $O$. Its circle passes through the three vertices, and its radius is a center-to-vertex distance.

Incenter $I$, incircle, inradius $r$

The internal angle bisectors meet at $I$. Its circle lies inside the triangle and touches each side once; its radius is the common perpendicular side distance.

Median and centroid $G$

A median joins a vertex to the opposite side midpoint. The three medians meet at $G$, which divides each median in the same ordered ratio.

Altitude, altitude line, orthocenter $H$

An altitude joins a vertex perpendicularly to the opposite supporting side line. The full altitude lines meet at $H$; their vertex-to-foot segments need not all contain $H$.

Uniform triangular sheet

A thin filled triangular region with mass distributed uniformly throughout, with no added weight. A wire boundary or a nonuniform sheet has a different balance model.

Formulas and theorems

• Perpendicular-bisector theorem and converse

  For distinct $A,B$: being on the perpendicular bisector implies $PA=PB$; conversely, equal endpoint distances put $P$ on that bisector.

  Use when The segment has positive length. The midpoint itself is included directly by its definition, even though connector triangles collapse there. Coincident endpoints instead give the whole plane as the equal-distance locus.

• Internal angle-bisector equidistance theorem

  A point on an internal angle bisector has equal perpendicular distances to the two supporting side lines.

  Use when The angle is strictly between zero and $180^\circ$. For a point other than the vertex, both perpendicular feet lie on the forward rays. At the vertex both line distances are zero.

• Interior-point converse

  An interior point with equal perpendicular distances to the two side lines lies on the internal angle-bisector ray.

  Use when The point is strictly inside the specified angle, whose measure is between zero and $180^\circ$. Equal distances to full lines alone also allow exterior possibilities; equal slanted distances do not meet the hypothesis.

• Circumcenter and circumcircle

  $$\begin{gathered} OA=OB\\ OB=OC\\ R=OA \end{gathered}$$

  Text description

  Acute triangle ABC has A and B along its lower side, with C above. M is the midpoint of AB and N is the midpoint of AC. Different tick patterns mark the two separate equal-half pairs. Dashed lines perpendicular to those sides at M and N intersect at O inside the triangle. The circle centered at O goes through all three vertices. The right-angle marks belong to the two side midpoints.

  Use when The vertices form a nondegenerate triangle. Two side perpendicular bisectors determine the unique center; the third is a consequence. The radius reaches a vertex, regardless of whether the center is inside.

• Incenter and incircle

  $$\begin{gathered} ID=IE\\ IE=IF\\ r=IF \end{gathered}$$

  Text description

  Triangle ABC is right at A, with AB horizontal and AC vertical. The internal angle bisectors from B and C meet at I inside the triangle. D, E and F are the perpendicular feet from I on BC, CA and AB. Right-angle squares identify perpendicular distances, and matching ticks identify their equal lengths. A circle centered at I passes through the three feet and stays inside the triangle. None of these feet is a triangle center name.

  Use when In nondegenerate $ABC$, $I$ is the internal-bisector intersection; $D,E,F$ are perpendicular feet on $BC,CA,AB$. Each foot lies strictly on its side segment. The circle reaches each supporting line only at its foot and fits inside the triangle.

• Centroid division of a median

  $$\begin{gathered} AG=2GD\\ AG=\frac23 AD\\ GD=\frac13 AD \end{gathered}$$

  Text description

  Scalene triangle ABC has side midpoints D on BC, E on CA, and F on AB. The three equal-half pairs use one, two, and three ticks respectively, so no equality between different sides is asserted. The medians AD, BE and CF meet at interior point G. On AD the order is A, G, D. The vertex portion AG is twice the midpoint portion GD. No right-angle mark or equal-angle mark is shown.

  Use when In a nondegenerate triangle, $D$ is the opposite side midpoint and $G$ is the centroid. The order is vertex $A$, then $G$, then midpoint $D$. A ratio on an arbitrary vertex-to-side segment is insufficient. If a median is also an altitude, its vertex lies on the opposite side’s perpendicular bisector, forcing equality of the two adjacent sides. One such median alone need not force an equilateral triangle.

• Altitude concurrence

  The three full altitude lines have exactly one common point, the orthocenter $H$.

  Text description

  Triangle ABC has an obtuse angle at A above base BC. The altitude from A reaches D inside BC. The altitude from B reaches E on line AC extended beyond A. The altitude from C reaches F on line AB extended beyond A. Right-angle squares mark D, E and F. The complete altitude lines, including the extensions past their feet where necessary, meet at H above the triangle. H is distinct from all three feet.

  Use when The triangle is nondegenerate. Each line passes through its own vertex and is perpendicular to the opposite supporting line. Extend sides and altitude lines as required; a foot and the common intersection are different objects.

• Center locations by angle type

  Acute: $O$ and $H$ are inside. Right: $O$ is the hypotenuse midpoint, while $H$ is the right-angle vertex. Obtuse: both are outside. The incenter $I$ and centroid $G$ remain strictly inside in all three cases.

  Use when For an obtuse angle at $C$, $O$ lies across the opposite side $AB$ from $C$. Location of one altitude foot alone does not determine $H$. These statements concern nondegenerate triangles.

• Equilateral coincidence and converse

  In a nondegenerate triangle, the centroid and orthocenter coincide exactly when the triangle is equilateral. In that case all four centers coincide.

  Use when In the equilateral case, each median line is also an internal angle bisector, altitude line, and perpendicular bisector. Conversely, $G$ is interior and distinct from each vertex, so a common $G=H$ makes every median line the corresponding altitude line. Each vertex then lies on the opposite side’s perpendicular bisector, giving adjacent-side equality at every vertex.

• Geometric centroid and physical balance

  A uniform thin triangular sheet balances at its centroid.

  Use when Material fills the triangular region uniformly. Equal outline or equal total mass alone does not guarantee the same balance point for a wire frame or weighted sheet. The geometric centroid still depends on the vertices.

Problem types, step by step

Exam traps

• Trap Treating any bisector or special segment as all the others

  Fix Require the actual midpoint, perpendicular, or equal-angle condition. A median need not be an altitude, and an altitude need not bisect the original side.

• Trap Measuring a line distance along a slanted segment

  Fix Use the perpendicular foot on the full line. If it misses a side segment or forward ray, that line distance is not the distance to the segment or ray.

• Trap Reversing equidistance without its hypotheses

  Fix Endpoint loci require distinct endpoints. The internal angle converse requires an interior point. Equal distances to three side lines identify the incenter only with interior location.

• Trap Using a vertex distance as the incircle radius

  Fix Circumradii reach vertices; inradii reach perpendicular feet. Keep $O,I,G,H$ for centers and give feet other names.

• Trap Putting the two-to-one ratio in the wrong order

  Fix The vertex portion is longer. The whole has three parts; the centroid is not the median midpoint. Prove the segment is a median before using the ratio.

• Trap Calling original altitude feet auxiliary midpoints

  Fix In the parallel-triangle proof, the original vertices are midpoints of auxiliary sides. Prove both midpoint placement and perpendicularity before invoking perpendicular-bisector concurrence.

• Trap Using a drawing or measurement as a concurrence proof

  Fix Recall the established theorem and its dependencies. A near crossing, equal measured lengths, or one interior half-plane does not establish the full claim.

• Trap Assuming all centers are interior or equal

  Fix Use the angle-type cases for $O,H$. Only $I,G$ are always interior. Equilateral coincidence requires the equilateral condition; a right triangle already separates $O$ from $H$.

• Trap Replacing uniform sheet by uniform wire

  Fix The balance conclusion requires uniform material throughout the filled triangle. A boundary frame, added weight, or uneven sheet changes the mass distribution.

• Trap Confusing the opposite-side ratio theorem with equidistance

  Fix The earlier angle-bisector ratio theorem divides the opposite side. Equidistance compares perpendicular distances to the two side lines.