Prove a Geometric Theorem

Prove geometric theorem rigorously: pick proof method, construct logical chain of justified steps from hypotheses → conclusion, handle special cases, produce complete proof doc.

Use When

• Geometric statement → prove true
• Verify conjecture about figures/relationships
• Establish lemma w/in larger argument
• Convert geometric insight → rigorous proof
• Compare proof method effectiveness for same theorem

In

• Required: Theorem statement (claim to prove)
• Required: Given info (hypotheses, defs, diagram desc)
• Optional: Preferred method (direct, contradiction, coordinate, vector, transformation)
• Optional: Rigor level (informal, semi-formal, formal w/ axiom citations)
• Optional: Known results citable w/o proof (e.g., "may assume Pythagorean")
• Optional: Address all degenerate + special cases explicitly?

Do

Step 1: State Theorem Precisely

Rewrite in standard form w/ explicit Given + Prove clauses.

1. Extract hypotheses. List every condition in "Given". Be explicit about geometric type (point, line, segment, ray, circle, polygon), incidence (lies on, passes through), metric (congruent, equal, perpendicular, parallel), ordering.

2. State conclusion. Exactly what must be proved in "Prove". Distinguish:

   • Equality/congruence: AB = CD, angle A = angle B, triangle ABC congruent to DEF
   • Incidence: P lies on L, three lines concurrent
   • Inequality: AB > CD, angle A < 90°
   • Existence: there exists P such that...
   • Uniqueness: such P is unique

3. Implicit assumptions. Many problems assume Euclidean (parallel postulate), non-degeneracy (points not coincident, lines not concurrent unless stated), positive orientation. Make explicit.

4. Draw or describe configuration. Diagram provided → transcribe key features. None → construct:

Given: Triangle ABC with D the midpoint of BC, E the midpoint of AC.
       Line segment DE.
Prove: DE is parallel to AB and DE = AB/2.

Configuration:
  A is at the apex; B and C form the base.
  D is the midpoint of BC; E is the midpoint of AC.
  DE connects the two midpoints.

Implicit assumptions: Euclidean plane, A is not on line BC (non-degenerate triangle).

→ Precise unambiguous statement w/ Given + Prove, all implicit assumptions surfaced, clear configuration desc.

If err: vague statement ("medial triangle similar to original") → rewrite w/ explicit defs + quantifiers. Statement appears false → test specific example before proceeding. False theorem unprovable; instead find + state counterexample.

Step 2: Select Proof Method

Pick technique best suited to theorem's structure.

Available methods + when to use:

1. Direct (synthetic): Forward from hypotheses via Euclidean propositions + established theorems.

   • Best for: congruence/similarity, angle chasing, incidence theorems.
   • Tools: triangle congruence (SSS, SAS, ASA, AAS, HL), parallels (alternate interior, corresponding angles), circle theorems (inscribed angle, tangent-radius, power of point).

2. By contradiction: Assume negation of conclusion, derive contradiction.

   • Best for: uniqueness, impossibility, inequality where direct unclear.
   • Structure: "Assume, for contradiction, [negation]. Then... But contradicts [known fact]. Therefore conclusion holds."

3. Coordinate: Place figure in coordinates + use algebra.

   • Best for: midpoint/distance/slope, collinearity, parallelism, perpendicularity.
   • Setup: choose coords to min computation (vertex at origin, side along axis).

4. Vector: Geometric relationships as vector ops.

   • Best for: centroid/barycenter, parallelism (parallel vectors), perpendicularity (dot product = 0), area ratios.
   • Notation: position vectors relative to origin, or free vectors for translation-invariant.

5. Transformation: Apply geometric transformation (reflection, rotation, translation, dilation) mapping part to part.

   • Best for: symmetry-based, congruence via isometry, similarity via dilation.

Document choice:

Theorem: Midline theorem (DE || AB and DE = AB/2).
Method evaluation:
  - Direct: requires parallel line theory and similar triangles. Moderate.
  - Coordinate: place B at origin, C on x-axis. Short computation. Good.
  - Vector: express D, E as midpoints, compute DE vector. Elegant.
Selected method: Coordinate proof (for explicit computation).
Alternative: Vector proof (for elegance).

→ Named method w/ justification, optionally alternative approaches noted.

If err: chosen method impasse after Step 3 → switch to alt. Coordinate proofs always settle metric questions mechanically → reliable fallback. Contradiction's negation doesn't lead to useful intermediate → try direct.

Step 3: Construct Proof w/ Justified Steps

Build proof as sequence of logical steps, each justified by axiom, def, or established result.

Direct/synthetic:

Chain of implications, each step cites justification:

Proof:
1. Let M be the midpoint of AB.                    [Given]
2. Then AM = MB = AB/2.                            [Definition of midpoint]
3. In triangle ABC, since CM is a median,
   CM connects vertex C to midpoint M of AB.       [Definition of median]
4. Triangles ACM and BCM share side CM.            [Common side]
5. AM = MB.                                         [Step 2]
6. AC may or may not equal BC.                      [No assumption of isosceles]
...

Coordinate:

Set up coords, compute, interpret:

Proof (coordinate):
1. Place B at the origin (0, 0) and C at (2c, 0).  [Choice of coordinates]
2. Let A = (2a, 2b) for some a, b with b != 0.     [Non-degeneracy; factor of 2
                                                      simplifies midpoint computation]
3. D = midpoint of BC = ((0 + 2c)/2, 0) = (c, 0).  [Midpoint formula]
4. E = midpoint of AC = ((2a + 2c)/2, (2b + 0)/2)
     = (a + c, b).                                   [Midpoint formula]
5. Vector DE = E - D = (a + c - c, b - 0) = (a, b). [Vector subtraction]
6. Vector AB = B - A = (0 - 2a, 0 - 2b) = (-2a, -2b).
   So vector BA = (2a, 2b) = 2 * (a, b) = 2 * DE.  [Vector subtraction]
7. Since BA = 2 * DE, vectors DE and BA are parallel
   (scalar multiple) and |DE| = |BA|/2.             [Parallel vectors; magnitude]
8. Therefore DE || AB and DE = AB/2.                 [QED]

Vector:

Position vectors relative to origin:

Proof (vector):
Let position vectors of A, B, C be a, b, c respectively.
1. D = (b + c)/2.                                   [Midpoint of BC]
2. E = (a + c)/2.                                   [Midpoint of AC]
3. DE = E - D = (a + c)/2 - (b + c)/2 = (a - b)/2. [Vector subtraction]
4. AB = B - A = b - a.                               [Vector subtraction]
5. DE = -(1/2)(b - a) = (1/2)(a - b).
   So DE = -(1/2) * AB, meaning DE = (1/2) AB
   in magnitude with opposite direction
   (equivalently, DE || AB).                         [Scalar multiple => parallel]
6. |DE| = (1/2)|AB|, i.e., DE = AB/2.               [Magnitude of scalar multiple]
QED.

Proof structure reqs:

• Number every step
• Cite justification in brackets after each
• "Therefore" or "Hence" → mark logical conclusions
• Avoid gaps: step needs intermediate result → prove it or cite it

→ Complete proof, every step follows from prev + cited results, no unjustified claims.

If err: step unjustifiable → may be false. Test specific example. Holds numerically but no justification → may need intermediate lemma. State + prove separately, resume main. Approach stuck → return Step 2, pick different method.

Step 4: Handle Special Cases + Edge Conditions

Configurations where general argument might fail.

1. Degenerate cases. Proof holds when:

   • Triangle → line (collinear vertices)
   • Circle → point (radius zero) or line (radius infinity)
   • Two points coincide
   • Angle = 0 or π (straight angle)
   • Polygon non-convex or self-intersecting

2. Boundary cases. Extreme values:

   • Right angles in angle-dependent
   • Isosceles or equilateral specializations in triangle theorems
   • Tangent vs secant in circle theorems

3. Coordinate proofs → verify coord assignment doesn't lose generality:

   • Did placing point at origin exclude valid config?
   • Did assuming side on axis force special orientation?
   • Implicit sign assumptions (b > 0) excluding valid cases?

4. Document each special case w/ resolution:

Special cases:
- If A lies on BC (degenerate triangle): D = E = midpoint of BC,
  and DE has length 0 while AB/2 > 0 in general. But the theorem
  assumes a non-degenerate triangle (b != 0 in our coordinates), so
  this case is excluded by hypothesis.
- If triangle is isosceles with AB = AC: the proof applies without
  modification (no special property of isosceles triangles was excluded).
- Coordinate generality: A = (2a, 2b) with b != 0 covers all non-degenerate
  triangles up to rotation and reflection, which preserves parallelism and
  length ratios. No generality lost.

→ Every degenerate/boundary case ID'd, each: proof applies unchanged, excluded by hypothesis, or separate argument provided.

If err: special case breaks proof → theorem may need additional hypothesis ("non-degenerate triangles"). Revise statement Step 1 or provide separate proof.

Step 5: Write Complete Proof w/ QED

Assemble final doc combining all elements.

1. Header: State theorem in Given/Prove form.

2. Proof body: Complete chain of justified steps from Step 3.

3. Special cases: From Step 4 either inline (brief) or remark after main.

4. Termination: Clear marker:

   • "QED" (quod erat demonstrandum)
   • Halmos tombstone (filled or open square)
   • "This completes the proof."

5. Review proof for logical completeness:

   • Every step from prev or cited results?
   • All hypotheses used? (unused → theorem holds under weaker conditions, or gap)
   • Conclusion reached explicitly in final step?

Format:

THEOREM (Midline Theorem):
Given: Triangle ABC; D is the midpoint of BC; E is the midpoint of AC.
Prove: DE || AB and DE = AB/2.

PROOF:
Place B = (0, 0), C = (2c, 0), A = (2a, 2b) with b != 0
(ensuring non-degeneracy).

(1) D = midpoint(B, C) = (c, 0).                 [Midpoint formula]
(2) E = midpoint(A, C) = (a + c, b).             [Midpoint formula]
(3) Vector DE = (a, b).                           [Subtraction: (2) - (1)]
(4) Vector BA = (2a, 2b) = 2 * DE.               [Subtraction: A - B]
(5) Since BA = 2 * DE, the vectors are parallel,
    so DE || AB.                                  [Parallel criterion]
(6) |DE| = sqrt(a^2 + b^2);
    |AB| = sqrt(4a^2 + 4b^2) = 2*sqrt(a^2 + b^2)
         = 2|DE|.
    Therefore DE = AB/2.                          [Magnitude computation]

QED.

6. Optional: state converse or note generalizations.

→ Self-contained proof doc readable hypothesis → conclusion w/o external refs, ending w/ explicit QED.

If err: review finds gap → return Step 3. Correct but excessively long (>30 steps) → restructure w/ lemmas: extract reusable intermediates, cite in main.

Check

• Theorem in precise Given/Prove form, all implicit assumptions explicit
• Proof method named + justified
• Every step numbered + cites justification
• No unjustified claims or logical gaps
• All hypotheses used ≥1 (or noted potentially removable)
• Conclusion explicit as final logical step
• Degenerate + boundary cases ID'd + addressed
• Coordinate proofs show choice loses no generality
• Proof ends w/ QED or equivalent
• Proof tested vs ≥1 specific numerical example

Traps

• Assume what you want to prove (circular reasoning): Most insidious error. Proving 2 triangles congruent → using consequence as step. Always trace each step back to hypotheses or established results, never to conclusion.

• Unjustified diagram assumptions: Diagram may suggest 2 lines intersect, point inside triangle, angle acute. Visual impressions must be proved, not assumed. Diagrams illustrate; not proof.

• Loss of generality in coord placement: Placing A at origin, B on positive x-axis, C in upper half-plane excludes clockwise vertex ordering. May not matter for distance/parallelism, but matters for orientation-dependent (signed area, cross product). Always verify.

• Overlook degenerate cases: Proof about triangles inscribed in circle may fail when triangle degenerates to diameter + point. Always check coincident points, parallel lines, degenerate figures.

• Cite more powerful result than needed: Law of cosines to prove what follows from basic angle-chasing → obscures logic + may add unnecessary assumptions (cosine well-defined). Use simplest sufficient tool.

• Miss the converse trap: "If quadrilateral is parallelogram, diagonals bisect each other" true, but converse = separate theorem needing separate proof. Don't prove converse when forward requested, or vice versa.

• Incomplete case analysis: Proof splits into cases (angle A acute, right, obtuse) → all cases addressed. Proving acute + claiming "others similar" w/o verification hides genuine differences.

→

• construct-geometric-figure — constructions + proofs complementary: constructions demonstrate existence, proofs establish properties
• solve-trigonometric-problem — trig computations often appear as sub-tasks in geometric proofs
• create-skill — packaging new proof technique as reusable skill