Prove and apply theorems about triangles: the angle sum, base angles of isosceles triangles, the midsegment, and the concurrence of the medians

Learning goal: Prove and apply theorems about triangles, both formally and informally, using a variety of methods: the measures of the interior angles of a triangle sum to 180°; the base angles of an isosceles triangle are congruent; the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length; and the three medians of a triangle meet at a single point.

The angle sum. Theorem: the measures of the interior angles of a triangle sum to 180°. Two-column proof for triangle ABC: (1) Through C draw the line parallel to AB; reason, through a point not on a line there is exactly one parallel. (2) Label the angle on this line to the left of C as ∠1 and to the right as ∠3, with ∠C between them. (3) ∠1 ≅ ∠A; reason, alternate interior angles formed by transversal AC crossing the parallel lines are congruent. (4) ∠3 ≅ ∠B; reason, alternate interior angles formed by transversal BC. (5) m∠1 + m∠C + m∠3 = 180°; reason, the three angles together form a straight angle along the line through C. (6) m∠A + m∠C + m∠B = 180°; reason, substitution from (3) and (4). The parallel line is the whole trick: it copies the angles at A and B to the vertex C, where they line up with ∠C along a straight line. Applying it: a triangle with angles (2x)°, (3x + 10)° and (x + 20)° satisfies 6x + 30 = 180, so x = 25 and the angles are 50°, 85° and 45°, which add to 180°. A corollary worth proving as an exercise: an exterior angle of a triangle equals the sum of the two remote interior angles, because the exterior angle and its adjacent interior angle sum to 180°, as do all three interior angles.

Base angles of an isosceles triangle. An isosceles triangle has two congruent sides, the legs; the third side is the base and the angles at its ends are the base angles. Theorem: the base angles of an isosceles triangle are congruent. Two-column proof for triangle ABC with AB = AC: (1) Let M be the midpoint of BC; reason, every segment has a midpoint. (2) Draw AM. (3) AB = AC; given. (4) BM = CM; reason, definition of midpoint. (5) AM = AM; reason, a segment is congruent to itself. (6) Triangle ABM ≅ triangle ACM; reason, SSS. (7) ∠B ≅ ∠C; reason, corresponding parts of congruent triangles are congruent. A paragraph proof by symmetry says the same thing with a rigid motion: reflect across the perpendicular bisector of BC, which passes through A because AB = AC (the perpendicular bisector theorem). The reflection swaps B and C, fixes A, and so carries ∠B onto ∠C; reflections preserve angle measure. Applying it: an isosceles triangle with apex angle 40° has base angles (180° − 40°) ÷ 2 = 70° each. The converse, that a triangle with two congruent angles is isosceles, is also true but is a separate theorem.

The midsegment. A midsegment joins the midpoints of two sides. Theorem: the midsegment is parallel to the third side and half its length. A coordinate proof handles every triangle at once by placing it generally: A(0, 0), B(2b, 0) and C(2c, 2d) with b > 0 and d not zero (the factor 2 is a convenience that keeps the midpoints free of fractions). The midpoint of AC is M(c, d) and the midpoint of BC is N(b + c, d). M and N have the same y-coordinate, so MN is horizontal, and AB lies on the x-axis, also horizontal: MN ∥ AB. The length MN = (b + c) − c = b, and AB = 2b, so MN is half of AB. Every step is arithmetic, and because b, c and d were arbitrary the proof covers all triangles. A concrete case: A(0, 0), B(8, 0), C(2, 6) has midpoints M(1, 3) and N(5, 3); MN is horizontal of length 4, and AB has length 8.

The medians meet at a point. A median joins a vertex to the midpoint of the opposite side. Theorem: the three medians of a triangle meet at one point, the centroid. Coordinate proof: place A(0, 0), B(6b, 0), C(6c, 6d) with b > 0 and d ≠ 0 (multiples of 6 keep every later coordinate whole). The midpoints are L(3b, 0) on AB, M(3c, 3d) on AC and N(3b + 3c, 3d) on BC. Let G be the point (2b + 2c, 2d). G is on median AN because it is two thirds of the way from A to N: (2/3)(3b + 3c, 3d) = (2b + 2c, 2d). G is on median BM because two thirds of the way from B to M is (6b, 0) + (2/3)((3c, 3d) − (6b, 0)) = (6b + 2c − 4b, 2d) = (2b + 2c, 2d). G is on median CL because two thirds of the way from C to L is (6c, 6d) + (2/3)((3b, 0) − (6c, 6d)) = (6c + 2b − 4c, 6d − 4d) = (2b + 2c, 2d). The same point lies on all three medians, so they are concurrent, and as a bonus the centroid divides each median in the ratio 2 : 1 from the vertex. Concrete case: A(0, 0), B(8, 0), C(2, 6) has centroid ((0 + 8 + 2) ÷ 3, (0 + 0 + 6) ÷ 3) = (10/3, 2); the median from A to N(5, 3) is the line y = (3/5)x, and at x = 10/3 it gives y = 2, so G is on it; the median from B(8, 0) to M(1, 3) has slope −3/7, and at x = 10/3 it gives y = (−3/7)(10/3 − 8) = (−3/7)(−14/3) = 2; the median from C(2, 6) to L(4, 0) has slope −3, and at x = 10/3 it gives y = 6 − 3(4/3) = 2. All three pass through (10/3, 2).

Choosing a method. The angle-sum and isosceles proofs are synthetic: they use congruent triangles and parallel lines. The midsegment and median proofs are coordinate proofs: they place the figure on axes with letters for the coordinates and compute. A transformational proof, like the reflection for the isosceles triangle, uses a rigid motion. Any of these is a complete proof when every step has a reason; an informal check, such as measuring the medians on one drawing, supports the theorem but does not prove it, because it covers only the triangle you drew. Ohio groups these theorems under a cluster that asks for proofs "both formally and informally using a variety of methods", and the honest habit is to say which one you are giving.

The misconception. Students often assume the coordinate proof only works for the triangle whose numbers they chose. It works for every triangle exactly when the coordinates are letters that could be anything, which is why the general proofs above use b, c and d rather than 8, 2 and 6. Placing A at the origin and B on the x-axis loses no generality, because any triangle can be moved there by a rigid motion, and rigid motions preserve every length and angle the theorem is about.

Worked, applying the angle sum and its corollary: A triangle has angles measuring (4x − 5)°, (3x + 15)° and (2x + 8)°. Their sum is 9x + 18 = 180, so x = 18 and the angles are 67°, 69° and 44°; check 67 + 69 + 44 = 180. The exterior angle at the 44° vertex equals 180° − 44° = 136°, and the two remote interior angles sum to 67 + 69 = 136°, confirming the corollary that an exterior angle equals the sum of the remote interior angles.

Worked, isosceles triangle with algebra: In triangle PQR, PQ = PR, m∠Q = (5x − 7)° and m∠R = (3x + 21)°. Base angles of an isosceles triangle are congruent, so 5x − 7 = 3x + 21, x = 14, and each base angle is 63°. The apex angle is 180° − 2(63°) = 54°. If instead the apex angle were given as 54°, the base angles would be (180° − 54°) ÷ 2 = 63° each, the same triangle from the other direction.

Worked, midsegment in a specific triangle and by the general proof: Triangle with A(0, 0), B(8, 0), C(2, 6). Midpoint of AC: M(1, 3). Midpoint of BC: N(5, 3). MN has slope 0 and length 5 − 1 = 4; AB has slope 0 and length 8. So MN ∥ AB and MN = ½ AB. The third midpoint, L(4, 0) on AB, gives two more midsegments: LM from (4, 0) to (1, 3) has run −3 and rise 3, while side BC from (8, 0) to (2, 6) has run −6 and rise 6, so LM ∥ BC with length √18 = ½ √72 = ½ BC; and LN from (4, 0) to (5, 3) has run 1 and rise 3, while AC has run 2 and rise 6, so LN ∥ AC with length √10 = ½ √40 = ½ AC. The three midsegments cut the triangle into four congruent triangles.

Worked, verifying the centroid and the 2 : 1 ratio: For A(0, 0), B(8, 0), C(2, 6), G = (10/3, 2). Distance from A to G is √((10/3)² + 4) = √(100/9 + 36/9) = √(136/9); distance from A to N(5, 3) is √(25 + 9) = √34 = √(306/9). The ratio AG ÷ AN = √(136 ÷ 306) = √(4/9) = 2/3, so AG : GN = 2 : 1. Along the median from C(2, 6) to L(4, 0), the point two thirds of the way is (2, 6) + (2/3)(2, −6) = (2 + 4/3, 6 − 4) = (10/3, 2) = G again.

Recording task: Draw triangle ABC with A(0, 0), B(8, 0), C(2, 6) on grid paper. Mark the midpoints L(4, 0), M(1, 3), N(5, 3); draw the midsegment MN and the three medians AN, BM, CL; and mark the centroid G(10/3, 2). Record: the slope and length of MN and of AB (both slope 0; lengths 4 and 8); the equation of each median and the check that (10/3, 2) satisfies it; the measures of the three angles by protractor and their sum. Then cut out a paper copy of the triangle and balance it on a pencil point at G, recording whether it balances, and write one sentence on why a balance demonstration is informal evidence rather than a proof.

Independent performance: Write a formal proof of one theorem from this lesson in a method different from the one shown for it (for example, a coordinate proof of the isosceles base-angle theorem, or a synthetic proof of the midsegment theorem using a parallelogram), and an informal demonstration of a second theorem using paper folding, tracing or software. Then solve two application problems you write yourself with algebraic angle measures. Success criteria: the formal proof places the figure generally (letters, not numbers) if it is a coordinate proof, or gives a reason for every statement if it is synthetic; the informal demonstration states clearly what it shows and why it is not a proof; each application names the theorem before the equation and checks the answer by substitution; and a closing paragraph explains why placing a triangle with one vertex at the origin and one side on an axis loses no generality.