Every triangle hides a small cast of special points: a point that is the same distance from all three corners, another that is the same distance from all three sides, a balance point, and a meeting place of the heights. In this chapter you will learn what these points are, why they exist, and how to use the segments behind them to find lengths, compare sides, and decide whether a triangle can even be built.
1. Midsegments of a triangle
A midsegment of a triangle is a segment that joins the midpoints of two sides. Every triangle has three midsegments.
A midsegment of a triangle is parallel to the third side and its length is half the length of that side. In the figure, \(\overline{DE}\parallel\overline{BC}\) and \(DE=\dfrac{1}{2}BC\).
This is a fast way to get a length without measuring. It also works backward: if you know the midsegment, double it to get the third side.
In \(\triangle ABC\), \(D\) and \(E\) are the midpoints of \(\overline{AB}\) and \(\overline{AC}\). Suppose \(DE=3x+1\) and \(BC=8x-6\), both in feet. Since \(BC=2\cdot DE\), we write \(8x-6=2(3x+1)=6x+2\), so \(2x=8\) and \(x=4\). Then \(DE=13\text{ ft}\) and \(BC=26\text{ ft}\). Check: \(2\cdot 13=26\).
2. Perpendicular bisectors and the circumcenter
A point lies on the perpendicular bisector of a segment if and only if it is the same distance from the two endpoints of the segment.
A triangle has three sides, so it has three perpendicular bisectors. They are always concurrent, meaning they meet at one point. That point is the circumcenter. Because it sits on all three bisectors, it is equidistant from the three vertices, so it is the center of the circumscribed circle that passes through \(A\), \(B\) and \(C\).
Where is the circumcenter? In an acute triangle it is inside. In a right triangle it is exactly the midpoint of the hypotenuse. In an obtuse triangle it falls outside the triangle.
Let \(A(0,0)\), \(B(10,0)\), \(C(4,8)\). The perpendicular bisector of \(\overline{AB}\) is the vertical line \(x=5\). The midpoint of \(\overline{AC}\) is \((2,4)\) and the slope of \(\overline{AC}\) is \(2\), so its bisector has slope \(-\dfrac{1}{2}\): \(y-4=-\dfrac{1}{2}(x-2)\). At \(x=5\) we get \(y=4-1.5=2.5\). The circumcenter is \(O(5,\,2.5)\). The radius is \(OA=\sqrt{25+6.25}=\sqrt{31.25}\approx 5.59\), and \(OC=\sqrt{1+30.25}\) gives the same value.
3. Angle bisectors and the incenter
A point lies on the bisector of an angle if and only if it is the same distance from the two sides of the angle. Distance always means the length of the perpendicular segment.
The three angle bisectors of a triangle meet at the incenter. It is equidistant from the three sides, so it is the center of the inscribed circle that touches each side once. Unlike the circumcenter, the incenter is always inside the triangle.
- Compute the semiperimeter \(s=\dfrac{a+b+c}{2}\).
- Find the area \(K\) of the triangle (for example with Heron’s formula \(K=\sqrt{s(s-a)(s-b)(s-c)}\)).
- The inradius is \(r=\dfrac{K}{s}\).
Here \(s=\dfrac{13+14+15}{2}=21\) and \(K=\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84\). So \(r=\dfrac{84}{21}=4\). The inscribed circle has radius 4 units.
The circumcenter uses perpendicular bisectors (they cut a side in half at a right angle), while the incenter uses angle bisectors. Do not mix them up: one is about distances to vertices, the other about distances to sides.
4. Medians and the centroid
A median of a triangle is a segment from a vertex to the midpoint of the opposite side.
The three medians meet at the centroid, the balance point of a flat triangular plate. It is always inside the triangle.
The centroid is \(\dfrac{2}{3}\) of the way from each vertex to the midpoint of the opposite side. If \(G\) is the centroid and \(M\) the midpoint of \(\overline{BC}\), then \(AG=2\cdot GM\), so \(AG=\dfrac{2}{3}AM\) and \(GM=\dfrac{1}{3}AM\).
With coordinates, \(G=\left(\dfrac{x_A+x_B+x_C}{3},\ \dfrac{y_A+y_B+y_C}{3}\right)\).
For \(A(2,1)\), \(B(8,7)\), \(C(5,10)\): \(G=\left(\dfrac{2+8+5}{3},\dfrac{1+7+10}{3}\right)=(5,6)\). Also, if a median is 24 cm long, the centroid is \(\dfrac{2}{3}\cdot 24=16\text{ cm}\) from the vertex and \(8\text{ cm}\) from the midpoint.
On my home planet we balance a triangular solar sail on a single pin. The pin goes exactly at the centroid, two thirds of the way along every median. Remember “2 from the corner, 1 from the side.”
5. Altitudes and the orthocenter
An altitude of a triangle is the perpendicular segment from a vertex to the line containing the opposite side. Its length is the height used in the area formula \(K=\dfrac{1}{2}bh\).
The three altitudes (or the lines that contain them) meet at the orthocenter. In an acute triangle it is inside, in a right triangle it is the vertex of the right angle, and in an obtuse triangle it is outside, where the extended altitude lines cross.
Take \(A(0,0)\), \(B(12,0)\), \(C(3,9)\). The altitude from \(C\) is the vertical line \(x=3\). The slope of \(\overline{BC}\) is \(\dfrac{9-0}{3-12}=-1\), so the altitude from \(A\) has slope \(1\): \(y=x\). They meet at \(H(3,3)\). Check with the third altitude: the slope of \(\overline{AC}\) is \(3\), so the altitude from \(B\) is \(y=-\dfrac{1}{3}(x-12)\), and at \(x=3\) it gives \(y=3\).
Here is a summary of the four centers. For any triangle, the circumcenter, centroid and orthocenter even lie on one line, called the Euler line.
| Center | Made from | Key property | Location |
|---|---|---|---|
| Circumcenter | perpendicular bisectors | equidistant from the vertices | inside, on the hypotenuse, or outside |
| Incenter | angle bisectors | equidistant from the sides | always inside |
| Centroid | medians | divides each median 2 : 1 | always inside |
| Orthocenter | altitudes | meeting point of the heights | inside, at the right-angle vertex, or outside |
6. Inequalities in one triangle
In a triangle, the longest side is opposite the largest angle and the shortest side is opposite the smallest angle. Conversely, the largest angle is opposite the longest side. Equal sides face equal angles.
In \(\triangle XYZ\), \(\angle X=52^\circ\), \(\angle Y=61^\circ\) and \(\angle Z=67^\circ\) (they add up to \(180^\circ\)). The side opposite \(\angle X\) is \(\overline{YZ}\), opposite \(\angle Y\) is \(\overline{XZ}\), and opposite \(\angle Z\) is \(\overline{XY}\). So \(YZ
7. The Triangle Inequality Theorem
The sum of the lengths of any two sides of a triangle is greater than the length of the third side. If two sides have lengths \(a\) and \(b\), the third side \(c\) must satisfy \(|a-b|
The idea is simple: the straight path from \(A\) to \(B\) is the shortest path. Going through a third point \(C\) can never be shorter, and it is strictly longer unless \(C\) lies on the segment.
- Order the three lengths and find the two smallest.
- Add the two smallest numbers.
- If the sum is strictly greater than the largest length, a triangle exists. If it is equal or smaller, it does not.
(a) Lengths 5 in, 9 in and 15 in: \(5+9=14\), and \(14<15\), so no triangle exists. (b) Two sides measure 8 cm and 15 cm. The third side \(x\) satisfies \(15-8
Key takeaways
- A midsegment joins two midpoints, is parallel to the third side, and measures half of it.
- Perpendicular bisectors meet at the circumcenter, which is equidistant from the three vertices.
- Angle bisectors meet at the incenter, which is always inside and equidistant from the three sides; \(r=\dfrac{K}{s}\).
- Medians meet at the centroid, which splits each median in a 2 : 1 ratio, vertex side first.
- Altitudes meet at the orthocenter, which is inside, on a vertex, or outside depending on the triangle type.
- The largest angle faces the longest side, and the smallest angle faces the shortest side.
- Any two sides must add up to more than the third side; the third side lies strictly between \(|a-b|\) and \(a+b\).
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