4.6 Concurrency
When three or more lines pass through a single point, we say that they are concurrent. In general, there is no reason to expect that three lines will all go through the same point. However, there are some special collections of lines that turn out to be concurrent. In this section, we look at some of these special cases. In the process, we will employ several of the important theorems we have encountered so far.
- Theorem 4.17: The perpendicular bisectors of the sides of a triangle are concurrent.
Given:
Lines
$\ell$, $m$ and $n$ are the perpendicular bisectors of sides
$\overline{AB}$, $\overline{BC}$, and $\overline{AC}$, respectively, of
$\triangle ABC$. Prove: Lines $\ell$, $m$, and $n$ are concurrent. |
Proof: The proof makes use of the Perpendicular Bisector Characterization Theorem. Since $\overline{AB}$ and $\overline{BC}$ are not collinear, their perpendicular bisectors are not parallel, so line $\ell$ intersects line $m$ at some point .
Since $P$ is on $\ell$, we know that $P$ is equidistant from points $A$ and $B$. Since $P$ is on $m$, we know that $P$ is equidistant from $B$ and $C$. Hence $PA=PB$ and $PB=PC$. That means that $PA=PC$, so that $P$ is equidistant from $A$ and $C$. But every point that is equidistant from $A$ and $C$ lies on the perpendicular bisector of $\overline{AC}$, so $P$ must lie on line $n$. Thus the three perpendicular bisectors of the sides of $\triangle ABC$ all pass through point $P$. They are concurrent.
The important consequence of this theorem is not so much that those three perpendicular bisectors are concurrent, but rather that their point of concurrency (point $P$ in the proof) is equidistant from all three vertices of the triangle. That means that if a circle is drawn with its center at $P$ and with PA as its radius, that circle will pass through all three vertices of the triangle — points $A$, $B$, and $C$. Since point $P$ is the center of a circle drawn around the triangle, we call $P$ the circumcenter of the triangle.
We now look at the angle bisectors of the angles of a triangle. To establish their concurrency, we make use of the Angle Bisector Characterization Theorem in exactly the same way we used the Perpendicular Bisector Characterization theorem in the last proof.
- Theorem 4.18: The bisectors of the three angles of a triangle are concurrent.
Given:
Rays $\ell$, $m$, and $n$ bisect $\angle A$, $\angle B$, and $\angle
C$, respectively, of $\triangle ABC$. Prove: Rays $\ell$, $m$, and $n$ are concurrent. |
Proof: We work in much the same way as we did in the last theorem. The bisectors of $\angle A$and $\angle B$ are not parallel, so they intersect at some point $P$. Since $P$ is on the bisector of $\angle CAB$, $P$ is equidistant from the sides of the angle, $\overline{AC}$ and $\overline{AB}$. Since $P$ is on the bisector of $\angle CBA$, $P$ is equidistant from the sides of that angle, $\overline{BA}$ and $\overline{BC}$. That means that $P$ is equidistant from all three sides of the triangle, so $P$ is equidistant from $\overline{CA}$ and $overline{CB}$. Then $P$ is on the bisector of $\angle ACB$, so all three angle bisectors go through $P$. They are concurrent.
What's not clear here is the significance of the new point $P$. It is equidistant from the
sides of $\triangle ABC$. Those distances are measured along perpendiculars to the sides of the angles. If one uses that distance as the radius of a circle, and the point $P$ as the center of the circle, one obtains a circle that is tightly enclosed within $\triangle ABC$. This is the inscribed circle of the triangle, and so point $P$ is called the incenter of the triangle.
We now turn to the altitudes of a triangle. To prove that they are concurrent, we need the theorem about the concurrency of the perpendicular bisectors of the sides of a triangle.
- Theorem 4.19: The lines containing the altitudes of a triangle are concurrent.
Given: The altitudes
of $\triangle ABC$ are $\overline{AP}$, $\overline{BQ}$, and
$\overline{CR}$. Prove: $\overleftrightarrow{AP}$, $\overleftrightarrow{BQ}$,and $\overleftrightarrow{CR}$ are concurrent. Construction: Through $A$, $B$, and $C$ draw the lines that are parallel to $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$, intersection at $X$, $Y$, and $Z$, as shown. |
Proof: Because of the parallelism, quadrilaterals $ABXC$, $ABCY$, and $AZBC$ are all parallelograms. Their opposite sides are congruent, so $\overline{CX}\cong\overline{AB}\cong\overline{CY}$, $\overline{AY}\cong\overline{BC}\cong\overline{AZ}$, and $\overline{BX}\cong\overline{AC}\cong\overline{BZ}$. The consequence of this is that $A$, $B$, and $C$ are the midpoints of $\overline{YZ}$, $\overline{XZ}$, and $\overline{XY}$. We know that, in a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other as well. That means that each of the altitudes of $\triangle ABC$ is perpendicular to one of the sides of $\triangle XYZ$. In particular, $\overleftrightarrow{AP}$ is the perpendicular bisector of $\overline{YZ}$, $\overleftrightarrow{BQ}$ is the perpendicular bisector of $\overline{XZ}$, and $\overleftrightarrow{CR}$ is the perpendicular bisector of $\overline{XY}$. Since we proved (in Theorem 4.17) that the perpendicular bisectors of the sides of a triangle are concurrent, we know that the altitudes $\overleftrightarrow{AP}$, $\overleftrightarrow{BQ}$, and $\overleftrightarrow{CR} of $\triangle ABC$ are concurrent.
- Theorem 4.20: The medians of a triangle are concurrent, with the point of concurrency two-thirds of the way along each one.
Our approach to this theorem is somewhat different from that taken in the first three concurrency problems. We start with two medians and show that the line from the third vertex through their point of intersection is, in fact, the third median. In the process we will show that the point of concurrency is two-thirds of the way along that third median, measuring from a vertex to the opposite side. Since that is true for the third median, it must, in fact, be true for the other two as well.
Given: $\overline{AM}$ and $\overline{BN}$ are medians of $\triangle ABC$, intersecting at $G$. Prove: $\overline{CG}$ passes through the midpoint of $\overline{AB}$ (so that the third median, drawn from $C$, passes through $G$ as well). Construction: Draw $\overline{CG}$, and extend it its own length to $F$, intersecting $\overline{AB}$ at $P$. Draw $\overline{FA}$ and $\overline{FB}$. |
Proof: By the construction, point $G$ is the midpoint of $\overline{CF}$. Since $\overline{AM}$ and $\overline{BN}$ are medians, $M$ is the midpoint of $\overline{BC}$ and $N$ is the midpoint of $\overline{AC}$. By the Midline Theorem, $\overline{GM}\parallel\overline{BF}$ and $\overline{GN}\parallel\overline{AF}$. But that means that $\overline{AG}\parallel\overline{BF}$ and $\overline{BG}\parallel\overline{AF}$, making quadrilateral $AFBG$ a parallelogram by definition. Since the diagonals of a parallelogram bisect each other, point $P$ is the midpoint of $\overline{AB}$, which means that $\overline{CGP}$ is, indeed, a median. Hence the three medians all intersect at point $G$. They are concurrent.
In addition, $CG=GF$ and $GF=2GP$, so $CG=2GP$. Then $CP=3GP$, and $CG=\frac{2}{3}CP$. That means that the point of concurrency is two-thirds of the way along median $\overline{CP}$.
We have mentioned the special names for the first two points of concurrency developed in this section — the circumcenter and the incenter of the triangle. Using the Greek prefix "ortho-," meaning "straight," "upright," or "erect," the point of concurrency of the lines containing the altitudes of a triangle is the orthocenter of the triangle. And, finally, since half of a triangle's area lies on each side of each of the three medians, the point of concurrency of the medians of a triangle is the centroid (or center of gravity) of the triangle.