Honors Geometry Unit 5 Review: Relationships in Triangles

Triangles have several special segments and lines, and each follows a different rule. A median runs from a vertex to the midpoint of the opposite side. An altitude runs from a vertex and meets the line containing the opposite side at a right angle. An angle bisector splits a vertex angle into two congruent angles. A perpendicular bisector is a line that crosses a side at its midpoint at a right angle, and it doesn't have to pass through any vertex.

Medians and altitudes get mixed up most often. A median is defined by a midpoint, and an altitude is defined by a right angle. They only coincide in special cases, like the segment from the vertex angle of an isosceles triangle to its base. In an obtuse triangle, two altitudes fall outside, so you extend the opposite side to find where they land.

Two equidistance facts come up in later proofs. Any point on a perpendicular bisector is the same distance from the segment's endpoints. Any point on an angle bisector is the same distance from the angle's two sides.

The Angle Bisector Theorem adds a proportion. If $\overline{AD}$ bisects $\angle A$ with $D$ on $\overline{BC}$, then $\frac{BD}{DC} = \frac{AB}{AC}$.

Worked example. In $\triangle ABC$, $AB = 6$, $AC = 9$, and $BC = 10$. The bisector of $\angle A$ meets $\overline{BC}$ at $D$. Let $BD = x$, so $DC = 10 - x$. Then $\frac{x}{10 - x} = \frac{6}{9}$, which gives $9x = 60 - 6x$, so $15x = 60$ and $x = 4$. That makes $BD = 4$ and $DC = 6$.

The Triangle Sum Theorem says the three interior angles of any triangle add to 180°. Almost every angle problem in this unit starts there. If two angles measure 50° and 70°, the third measures 180° − 120° = 60°.

An exterior angle forms when you extend one side past a vertex. The Exterior Angle Theorem says it equals the sum of the two remote interior angles, which are the two angles not touching it. It also forms a linear pair with the interior angle next to it, so those two add to 180°. Both facts come from the Triangle Sum Theorem, so if you forget one you can rebuild it from the other.

Isosceles triangles give you another tool. The Base Angles Theorem says the angles opposite congruent sides are congruent, and the converse is also true. Each angle of an equilateral triangle measures 60°.

Honors proofs often draw an auxiliary line through one vertex parallel to the opposite side. Alternate interior angles then show the triangle's three angles fit together along a straight line, which proves the 180° sum.

Worked example. An exterior angle measures $(5x + 10)$°, and its remote interior angles measure $(2x + 20)$° and $(x + 40)$°. Set $5x + 10 = (2x + 20) + (x + 40) = 3x + 60$. Then $2x = 50$, so $x = 25$. The exterior angle is 135°, the remote angles are 70° and 65°, and the adjacent interior angle is 180° − 135° = 45°. As a check, 70 + 65 + 45 = 180.

The Triangle Inequality Theorem says any two sides of a triangle must add to more than the third side, and the sum has to be strictly greater, not equal. Lengths 3, 4, and 7 fail because 3 + 4 = 7, so those segments would lie flat along a line.

You don't have to test all three sums. Add the two shortest sides and compare with the longest. If that sum is bigger, the other two checks pass automatically.

When you know two sides $a$ and $b$, the third side $x$ falls in a range given by $|a - b| < x < a + b$.

Sides and angles are ordered together. The longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. In a triangle with angles of 40°, 60°, and 80°, the side across from the 80° angle is longest.

An indirect proof assumes the opposite of what you want to prove, reasons forward, and reaches a contradiction. To show a triangle can't have two right angles, assume it does. Those two angles already total 180°, leaving 0° for the third angle, which is impossible, so the assumption was false.

Worked example. Two sides of a triangle are 5 and 8. The difference is $|5 - 8| = 3$ and the sum is $5 + 8 = 13$, so the third side satisfies $3 < x < 13$. If $x$ has to be a whole number, it can be any value from 4 through 12, which is 9 possible lengths.

Each family of special lines meets at a single point, called a point of concurrency. The lines containing the angle bisectors meet at the incenter, perpendicular bisectors at the circumcenter, lines containing the medians at the centroid, and lines containing the altitudes at the orthocenter. To find a center, you only need two of the lines, since the third always passes through the same point.

Connect each center to what it's equidistant from. The incenter is the same distance from all three sides, so it's the center of the inscribed circle. The circumcenter is the same distance from all three vertices, so it's the center of the circle passing through every corner.

The centroid splits every median in a 2:1 ratio, with the longer piece on the vertex side. It's the balance point of a triangular cutout, and on a coordinate plane it's the average of the three vertices.

Location depends on triangle type. The incenter and centroid always sit inside. The circumcenter and orthocenter are inside for acute triangles and outside for obtuse ones. In a right triangle, the circumcenter is the midpoint of the hypotenuse and the orthocenter is the right-angle vertex. In a triangle that is not equilateral, the orthocenter, centroid, and circumcenter lie on the Euler line. In an equilateral triangle, the three centers coincide, so there is no unique Euler line.

Worked example. A triangle has vertices $A(0, 0)$, $B(6, 0)$, and $C(3, 9)$. The centroid is $G = \left(\frac{0+6+3}{3}, \frac{0+0+9}{3}\right) = (3, 3)$. The median from $C$ ends at the midpoint of $\overline{AB}$, which is $(3, 0)$, so it has length 9. Then $CG = 6$ and the rest is 3, matching the 2:1 ratio.