Grade 7

Triangles and their special lines

The simplest polygon $\triangle ABC$ , together with its medians, angle bisectors ( $\dfrac{BD}{DC} = \dfrac{AB}{AC}$ ), altitudes, and perpendicular bisectors.

IntuitionThree vertices, four remarkable meeting points

Take any triangle $\triangle ABC$ on a sheet of paper and draw the three medians (each joining a vertex to the midpoint of the opposite side). No matter how lopsided the triangle is, all three lines pass through a single point $G$ , the centroid — the exact balance point where a cardboard cutout of $\triangle ABC$ would sit level on a pin! The same miracle happens three more times: the three angle bisectors meet at $I$ (the center of the inscribed circle), the three altitudes meet at $H$ (the orthocenter), and the three perpendicular bisectors meet at $O$ (the center of the circumscribed circle). The network diagram below shows the three vertices of $\triangle ABC$ together with these concurrent lines, a preview of the geometric structure we will prove below.

A triangle inscribed in a circle of radius

a

, showing its circumcenter

O

(green), centroid

G

(amber), and orthocenter

H

(red) collinear along the Euler line.

SchoolDefinitions and the four special lines

Definition: Angle sum and the four special lines of a triangle

In any triangle $\triangle ABC$ in the Euclidean plane, the three interior angles always add up to a straight angle:

\angle A + \angle B + \angle C = 180^\circ

The angle bisector $AD$ of $A$ (with $D$ on $BC$ ) divides the opposite side in the exact ratio of the two adjacent sides, and the three medians meet at the centroid $G$ two-thirds of the way along each median:

\frac{BD}{DC} = \frac{AB}{AC}

G = \left(\frac{x_A + x_B + x_C}{3},\, \frac{y_A + y_B + y_C}{3}\right), \qquad AG = \frac{2}{3} AM

The four special lines and their concurrency points
Line	Definition	Concurrency point	Key property
Median	vertex to opposite midpoint	Centroid $G$	$AG = \dfrac{2}{3} AM$ (divides in ratio $2:1$ )
Angle bisector	splits vertex angle in half	Incenter $I$	equidistant from all three sides; $\dfrac{BD}{DC} = \dfrac{AB}{AC}$
Altitude	vertex perpendicular to opposite side	Orthocenter $H$	lies inside iff triangle is acute
Perpendicular bisector	perpendicular at midpoint of side	Circumcenter $O$	equidistant from all three vertices

UndergraduateTwo key theorems and their proofs

Theorem: Angle Bisector Theorem

In $\triangle ABC$ , if the internal bisector of angle $A$ meets side $BC$ at $D$ , then $\dfrac{BD}{DC} = \dfrac{AB}{AC}$ .

Why is it true?

This converts information about angles (that $AD$ splits $A$ into two equal halves) directly into a ratio of lengths along the opposite side, letting you find where the bisector lands using only the three side lengths of $\triangle ABC$ .

Proof

Step 1 (compare areas using the shared altitude from $A$ ). Triangles $\triangle ABD$ and $\triangle ACD$ share the same altitude from vertex $A$ down to the line $BC$ , so their areas are proportional to their bases on $BC$ : $\dfrac{S(\triangle ABD)}{S(\triangle ACD)} = \dfrac{BD}{DC}$ .

Step 2 (compare the same two areas using $AB$ and $AC$ as bases). Because $D$ lies on the angle bisector of $A$ , the perpendicular distances from $D$ to the two sides of the angle are equal: $d(D, AB) = d(D, AC)$ . Taking $AB$ and $AC$ as the bases of $\triangle ABD$ and $\triangle ACD$ , those two equal perpendiculars are the corresponding altitudes, so the area ratio also equals the ratio of these bases: $\dfrac{S(\triangle ABD)}{S(\triangle ACD)} = \dfrac{AB}{AC}$ .

Step 3 (equate the two expressions). Since both right-hand sides equal the same area ratio, equating Step 1 and Step 2 immediately gives $\dfrac{BD}{DC} = \dfrac{AB}{AC}$ , completing the proof.

Theorem: The three medians concur at the centroid

In any triangle $\triangle ABC$ with medians $AM$ , $BN$ , $CP$ , all three medians pass through the single point $G = \left(\dfrac{x_A + x_B + x_C}{3},\, \dfrac{y_A + y_B + y_C}{3}\right)$ , which divides each median in the ratio $2:1$ from the vertex ( $AG = \dfrac{2}{3} AM$ ).

Why is it true?

Three random lines in a plane almost never pass through the same point — they form a small triangle instead. Seeing that the formula for the point $\dfrac{2}{3}$ of the way along $AM$ comes out completely symmetric in $A$ , $B$ , $C$ explains in one stroke why all three medians must hit the exact same spot.

Proof

Step 1 (midpoint coordinates). Place $\triangle ABC$ in a coordinate plane with vertices $A$ , $B$ , $C$ . The midpoint $M$ of side $BC$ is the average of $B$ and $C$ : $M = \left(\dfrac{x_B+x_C}{2},\, \dfrac{y_B+y_C}{2}\right)$ .

Step 2 (point two-thirds along $AM$ ). Move from $A$ toward $M$ by $\dfrac{2}{3}$ of the segment $AM$ : the resulting point is $A + \dfrac{2}{3}(M - A) = \dfrac{1}{3}A + \dfrac{2}{3}\cdot\dfrac{B+C}{2} = \dfrac{A+B+C}{3}$ , giving coordinates $G = \left(\dfrac{x_A + x_B + x_C}{3},\, \dfrac{y_A + y_B + y_C}{3}\right)$ .

Step 3 (symmetry forces concurrency). Notice that the final formula $G = \left(\dfrac{x_A + x_B + x_C}{3},\, \dfrac{y_A + y_B + y_C}{3}\right)$ is completely symmetric in $A$ , $B$ , $C$ — swapping the roles of $A$ and $B$ (to find the point $\dfrac{2}{3}$ of the way along median $BN$ ) or of $A$ and $C$ (along $CP$ ) produces the exact same point $G$ . Therefore all three medians pass through $G$ , and $G$ lies $\dfrac{2}{3}$ of the way from each vertex to the opposite midpoint ( $AG = \dfrac{2}{3} AM$ ).

UndergraduateReal-World Applications and Worked Examples

Triangle centers show up constantly in engineering and design: structural engineers and computer-graphics engines compute the center of mass of every triangular mesh face using the centroid formula $G = \left(\dfrac{x_A + x_B + x_C}{3},\, \dfrac{y_A + y_B + y_C}{3}\right)$ (a complex 3D model in a game or CAD program is made of thousands of flat triangles whose individual centroids are averaged, weighted by area, to find the object's balance point for physics simulation); surveyors and architects use the incenter $I$ to place the largest possible circular fountain or roundabout inside a triangular plot of land (since $I$ is equidistant from all three boundary roads), and the circumcenter $O$ to locate a cell tower or emergency beacon equidistant from three towns at $A$ , $B$ , $C$ .

Example: Balancing a triangular metal plate

A flat uniform triangular steel plate has corners at $A(0,6)$ , $B(3,0)$ , $C(9,3)$ (measured in decimeters on a workshop grid). At what coordinates $G$ should a single vertical support rod be welded underneath so the plate balances horizontally, and how far is $G$ from vertex $A$ along the median $AM$ ?

Solution

Step 1: apply the centroid formula. Averaging the three vertex coordinates from $A(0,6)$ , $B(3,0)$ , $C(9,3)$ gives $G = \left(\dfrac{x_A + x_B + x_C}{3},\, \dfrac{y_A + y_B + y_C}{3}\right)$ = $((0+3+9)/3,\, (6+0+3)/3)$ , which simplifies to $G(4,3)$ .

Step 2: find the distance from $A$ to $G$ . From $A(0,6)$ to $G(4,3)$ , the horizontal change is $4 - 0 = 4$ and the vertical change is $3 - 6 = -3$ , so by the Pythagorean theorem $AG = \sqrt{4^2 + (-3)^2} = 5$ dm.

Step 3: check using the midpoint $M$ . The midpoint of $BC$ is $M(6, 1.5)$ , whose distance from $A(0,6)$ is $AM = \sqrt{6^2 + (-4.5)^2} = 7.5$ dm, and indeed $AG = \dfrac{2}{3} AM$ gives $(2/3)\times 7.5 = 5$ dm — both methods agree.

Example: Dividing a side with an angle bisector

A triangular garden plot $\triangle ABC$ has side lengths $AB = 6$ , $AC = 9$ , $BC = 10$ (in meters). A straight path $AD$ is laid along the angle bisector of corner $A$ to meet the fence $BC$ at gate $D$ . Find the exact lengths of the two fence segments $BD$ and $DC$ .

Solution

Step 1: write the Angle Bisector Theorem ratio. By $\dfrac{BD}{DC} = \dfrac{AB}{AC}$ with $AB = 6$ , $AC = 9$ , $BC = 10$ and $DC = 10 - BD$ , we get $\dfrac{BD}{10 - BD} = \dfrac{6}{9} = \dfrac{2}{3}$ .

Step 2: solve the proportion for $BD$ . Cross-multiplying gives $3\,BD = 2(10 - BD) = 20 - 2\,BD$ , so $5\,BD = 20$ and $BD = 4$ m.

Step 3: find $DC$ and verify. Subtracting from the full side $BC = 10$ gives $BD = 4$ , $DC = 6$ (in meters), and indeed $BD/DC = 4/6 = 2/3 = 6/9 = AB/AC$ .

Which three special lines of $\triangle ABC$ intersect at the center $I$ of the inscribed circle (equidistant from all three sides)?

In $\triangle ABC$ , if median $AM$ has length $12$ cm, what is the distance $AG$ from vertex $A$ to the centroid $G$ ?

In $\triangle ABC$ with $AB = 4$ , $AC = 6$ , and angle bisector $AD$ meeting $BC$ at $D$ , what is the ratio $BD/DC$ ?

A cell tower must be built at equal distance from three villages located at the vertices $A$ , $B$ , $C$ of $\triangle ABC$ . Which triangle center gives the correct location?

References

H. S. M. Coxeter, S. L. Greitzer (1967). Geometry Revisited
Euclid (trans. Thomas L. Heath) (1956). Euclid's Elements (Books I–XIII)