Praxis Geometry (5163)

This chapter covers the triangle portion of outline area I.A: classifying triangles, the angle-sum and exterior-angle theorems, the triangle inequality, the special right triangles (45-45-90 and 30-60-90), and the special segments of a triangle (medians, altitudes, angle bisectors, perpendicular bisectors, and midsegments) together with their points of concurrency. These facts appear constantly in later sections, so aim for automatic recall.

By sides, a triangle is scalene (no congruent sides), isosceles (at least two congruent sides), or equilateral (three congruent sides). Under the inclusive definition used in most modern texts, every equilateral triangle is also isosceles. By angles, a triangle is acute (three acute angles), right (one right angle), or obtuse (one obtuse angle). An equiangular triangle has three 60° angles and is equilateral.

In an isosceles triangle, the two congruent sides are the legs, the third side is the base, and the angles opposite the legs are the base angles. The Isosceles Triangle Theorem says that if two sides of a triangle are congruent, the angles opposite them are congruent; its converse says that if two angles are congruent, the sides opposite them are congruent. In an isosceles triangle, the bisector of the vertex angle is also the median to the base, the altitude to the base, and part of the perpendicular bisector of the base.

You can classify a triangle by its side lengths using the converse of the Pythagorean theorem and its extensions. With c the longest side: if a^2 + b^2 = c^2, the triangle is right; if a^2 + b^2 > c^2, it is acute; if a^2 + b^2 < c^2, it is obtuse. For sides 6, 7, and 9: 36 + 49 = 85 and 81 < 85, so the triangle is acute.

The Triangle Angle-Sum Theorem states that the interior angles of a triangle add to 180°.

Proof: Given △ABC, prove that m∠A + m∠B + m∠C = 180°.

Through B, draw line l parallel to AC (Parallel Postulate: through a point not on a line there is exactly one parallel line)

Let ∠1 and ∠3 be the angles that l makes with BA and BC, with ∠1 next to BA and ∠3 next to BC, so that ∠1, ∠ABC, and ∠3 together form a straight angle at B (Angle Addition and definition of straight angle)

m∠1 + m∠ABC + m∠3 = 180° (Definition of a straight angle)

∠1 ≅ ∠A and ∠3 ≅ ∠C (Alternate interior angles formed by parallel lines are congruent)

m∠A + m∠ABC + m∠C = 180° (Substitution)

This completes the proof.

An exterior angle of a triangle is formed by one side and the extension of an adjacent side. The Exterior Angle Theorem says that an exterior angle equals the sum of the two remote (non-adjacent) interior angles. It follows from the angle sum: the exterior angle and its adjacent interior angle form a linear pair, so both the exterior angle and the sum of the remote angles equal 180° minus the adjacent angle. A corollary is that an exterior angle is greater than either remote interior angle. The exterior angles of a triangle, one at each vertex, add to 360°.

Worked Example: In △PQR, side QR is extended past R to point S. The exterior angle ∠PRS measures 125°, m∠P = (2x + 5)°, and m∠Q = (3x − 10)°. Find all three interior angles.

By the Exterior Angle Theorem, m∠P + m∠Q = m∠PRS: (2x + 5) + (3x − 10) = 125.

Simplify: 5x − 5 = 125, so 5x = 130 and x = 26.

∠PRQ and ∠PRS form a linear pair, so m∠PRQ = 180° − 125° = 55°.

Check: 57 + 68 + 55 = 180.

Result: The angles measure 57°, 68°, and 55°.

The Triangle Inequality Theorem states that the sum of any two side lengths of a triangle is greater than the third. In practice, if two sides are a and b with a ≤ b, the third side s must satisfy b − a < s < a + b. The inequality is strict: lengths 3, 4, and 7 do not form a triangle, because the "triangle" would collapse into a segment.

A companion result relates sides and angles: in any triangle, the longest side is opposite the largest angle and the shortest side is opposite the smallest angle. If m∠A = 50°, m∠B = 70°, and m∠C = 60°, then the sides from shortest to longest are BC, AB, AC (opposite A, C, and B respectively).

Worked Example: Two sides of a triangle measure 7 and 12. How many different integer lengths are possible for the third side?

The integers strictly between 5 and 19 are 6 through 18.

Two right triangles appear so often that you should know their side ratios cold. Answers are usually expected in exact form, such as 5 sqrt(3), rather than as decimals.

The 45-45-90 triangle is an isosceles right triangle, half of a square cut along a diagonal. If each leg is s, the hypotenuse is s sqrt(2) by the Pythagorean theorem (s^2 + s^2 = 2s^2). To go from the hypotenuse back to a leg, divide by sqrt(2): a hypotenuse of 10 gives legs of 10/sqrt(2) = 5 sqrt(2).

The 30-60-90 triangle is half of an equilateral triangle cut by an altitude. If the equilateral triangle has side 2s, the altitude splits the base into two segments of length s, so the short leg (opposite 30°) is s, the hypotenuse is 2s, and the long leg (opposite 60°) is sqrt((2s)^2 − s^2) = sqrt(3s^2) = s sqrt(3).

Special Right Triangle Ratios

Memory Aid: In the 30-60-90 triangle, the sides follow the angles: the smallest angle (30°) faces the shortest side s, the middle angle (60°) faces the middle side s sqrt(3), and the right angle faces the longest side 2s. Since sqrt(3) ≈ 1.73 lies between 1 and 2, the order always makes sense.

Worked Example: An equilateral triangle has side length 10. Find its height and its area, exactly and to the nearest hundredth.

The altitude bisects the base, creating two 30-60-90 triangles with hypotenuse 10 and short leg 5.

The altitude is the long leg: 5 sqrt(3) ≈ 8.66.

Check with the formula (a^2 sqrt(3))/4 = (100 sqrt(3))/4 = 25 sqrt(3).

Result: Height 5 sqrt(3) ≈ 8.66; area 25 sqrt(3) ≈ 43.30 square units.

Common Trap: Attaching sqrt(3) to the wrong leg. The long leg, opposite 60°, carries the sqrt(3). If the long leg is given (say 9), the short leg is 9/sqrt(3) = 3 sqrt(3), not 9 sqrt(3).

Every triangle has four families of special segments or lines, three of each, and each family meets at a single point. Knowing which point belongs to which family, and what property it has, is a frequent test target.

Points of Concurrency in a Triangle

A median joins a vertex to the midpoint of the opposite side. The three medians meet at the centroid, which lies two-thirds of the way from each vertex to the opposite midpoint, so it splits each median in a 2:1 ratio with the longer part next to the vertex. The centroid is the balance point (center of mass) of a uniform triangular region, and in coordinates it is the average of the vertices: G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3). Each median also splits the triangle into two triangles of equal area, and the three medians together split it into six small triangles of equal area.

An altitude is the perpendicular segment from a vertex to the line containing the opposite side. The lines containing the altitudes meet at the orthocenter. The orthocenter is inside an acute triangle, at the right-angle vertex of a right triangle, and outside an obtuse triangle.

An angle bisector divides an angle into two congruent angles. The three angle bisectors meet at the incenter, which is always inside the triangle and is equidistant from all three sides; that distance is the radius of the inscribed circle. A useful formula is r = Area/s, where s is the semiperimeter.

The perpendicular bisectors of the sides meet at the circumcenter, which is equidistant from all three vertices; that distance is the radius of the circumscribed circle. The circumcenter is inside an acute triangle, at the midpoint of the hypotenuse of a right triangle, and outside an obtuse triangle. So in any right triangle, the circumradius is half the hypotenuse.

For a median of length 18, the centroid is two-thirds of the way from the vertex: AG = (2/3)(18) = 12.

The remaining part is GD = 18 − 12 = 6, which confirms the ratio AG: GD = 12: 6 = 2: 1.

Result: G = (4, 5); for a median of 18, AG = 12 and GD = 6.

A midsegment of a triangle joins the midpoints of two sides. The Triangle Midsegment Theorem states that a midsegment is parallel to the third side and half as long. Each triangle has three midsegments, and they divide the triangle into four congruent triangles, each similar to the original with scale factor 1/2 (so each has one-fourth of the area). The midsegment triangle has half the perimeter of the original.

Result: x = 5, MN = 13, KL = 26.

For a right triangle with legs 6 and 8, the hypotenuse is 10. The circumcenter is the midpoint of the hypotenuse, so the circumradius is 5. The incenter radius is r = Area/s = 24/12 = 2 (the area is (1/2)(6)(8) = 24 and the semiperimeter is (6 + 8 + 10)/2 = 12). For right triangles there is also a shortcut: r = (a + b − c)/2 = (6 + 8 − 10)/2 = 2. The orthocenter is at the right-angle vertex. Questions that ask which center lies on the hypotenuse, or which lies at a vertex, test exactly these facts.

Expect numeric entry for angle measures, side lengths in special right triangles (often in exact radical form), and segment lengths along medians. Select-all items may ask which points of concurrency must lie inside a given type of triangle (only the centroid and incenter always do). Drop-down proof items may ask you to name a center from a description of how it was constructed. Use the on-screen calculator to check radicals, but enter the exact form when the question asks for it.

Interior angles of a triangle sum to 180°; an exterior angle equals the sum of the two remote interior angles.

The third side of a triangle lies strictly between the difference and the sum of the other two sides; the longest side is opposite the largest angle.

45-45-90 sides are s, s, s sqrt(2); 30-60-90 sides are s, s sqrt(3), 2s with sqrt(3) on the side opposite 60°.

Medians meet at the centroid (2:1 from the vertex, coordinates are the vertex averages); altitudes meet at the orthocenter; angle bisectors meet at the incenter (equidistant from sides); perpendicular bisectors meet at the circumcenter (equidistant from vertices).

In a right triangle, the circumcenter is the midpoint of the hypotenuse and the orthocenter is the right-angle vertex.

A midsegment is parallel to the third side and half its length.