executing-plans
executing-plans
Use when executing an implementation plan in the current session as the implementer yourself — your human partner chose inline execution, or no subagent tool is available
prove-geometric-theoremProve geometric theorems using Euclidean axiomatic methods, coordinate geometry, or vector methods with rigorous step-by-step logical structure. Covers direct proof, proof by contradiction, coordinate proofs, vector proofs, and handling of special cases and degenerate configurations. Use when given a geometric statement to prove, verifying a conjecture, est…
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Prove geometric theorem rigorously. Pick appropriate proof method. Build logical chain of justified steps from hypotheses to conclusion. Handle all special cases. Produce complete proof document.
Rewrite theorem in standard mathematical form with explicit Given and Prove clauses.
Extract hypotheses. List every condition in "Given" clause. Be explicit about geometric type (point, line, segment, ray, circle, polygon), incidence relations (lies on, passes through), metric conditions (congruent, equal, perpendicular, parallel), ordering assumptions.
State conclusion. Write exactly what must be proved in "Prove" clause. Distinguish between:
Identify implicit assumptions. Many geometry problems assume Euclidean geometry (parallel postulate), non-degeneracy (points not coincident, lines not concurrent unless stated), positive orientation. Make these explicit.
Draw or describe configuration. Diagram provided? Transcribe key features. If not, construct one:
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).
Got: Precise, unambiguous statement with Given and Prove clauses. All implicit assumptions surfaced. Clear description of geometric configuration.
If fail: Theorem statement vague (e.g. "the medial triangle is similar to the original")? Rewrite with explicit definitions and quantifiers. Statement appears false? Test with specific example before proceeding. False theorem can't be proved — instead, find and state counterexample.
Pick proof technique best suited to theorem's structure.
Available methods and when to use them:
Direct (synthetic) proof: Work forward from hypotheses using Euclidean propositions and previously established theorems.
Proof by contradiction: Assume negation of conclusion. Derive contradiction.
Coordinate proof: Place figure in coordinate system. Use algebra.
Vector proof: Express geometric relationships using vector operations.
Transformation proof: Apply geometric transformation (reflection, rotation, translation, dilation) that maps part of figure to another part.
Evaluate and 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).
Got: Named proof method with reason for why it suits this theorem. Optionally a note on alternative approaches.
If fail: First chosen method leads to impasse after Step 3? Switch to alternative. Coordinate proofs can always settle metric questions mechanically — serve as reliable fallback. Contradiction chosen but negation doesn't lead to useful intermediate statement? Try direct approach instead.
Build proof as sequence of logical steps. Each justified by axiom, definition, or previously established result.
For direct/synthetic proofs:
Organize as chain of implications. Each step must cite its 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]
...
For coordinate proofs:
Set up coordinates. 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]
For vector proofs:
Use position vectors relative to chosen 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 requirements:
Got: Complete proof where every step follows logically from previous steps and cited results. No unjustified claims.
If fail: Step can't be justified? May be false. Test with specific example. Holds numerically but you can't find justification? May require intermediate lemma. State lemma. Prove it separately. Resume main proof. Entire approach stuck? Return to Step 2 and pick different method.
Identify and address configurations where general argument might fail.
Degenerate cases. Check if proof holds when:
Boundary cases. Check extreme values:
For coordinate proofs, verify coordinate assignment doesn't lose generality:
Document each special case with its 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.
Got: Every degenerate or boundary case identified. For each: proof shown to apply unchanged, case shown to be excluded by hypothesis, or separate argument provided.
If fail: Special case breaks proof? Theorem may need additional hypothesis (e.g. "for non-degenerate triangles"). Revise theorem statement in Step 1 to include necessary condition. Or provide separate proof for special case.
Assemble final proof document combining all elements from previous steps.
Header: State theorem in Given/Prove form.
Proof body: Present complete chain of justified steps from Step 3.
Special cases: Include analysis from Step 4 either inline (if brief) or as remark after main proof.
Termination: End with clear marker:
Review proof for logical completeness:
Format final proof:
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.
Got: Self-contained proof document a reader (or verifying agent) can follow from hypothesis to conclusion without external references. Ends with explicit QED.
If fail: During final review gap found? Return to Step 3 to fill it. Proof correct but excessively long (over 30 steps)? Restructure with lemmas: extract reusable intermediate results as named lemmas proved separately, then cite them in main proof.
Assume what you want to prove (circular reasoning): Most insidious error. Example: in proving two triangles congruent, using consequence of that congruence as step. Always trace each step back to hypotheses or previously established results — never to conclusion.
Unjustified diagram assumptions: Diagram may suggest two lines intersect, point lies inside triangle, angle is acute. These visual impressions must be proved, not assumed. Diagrams illustrate. Don't constitute proof.
Loss of generality in coordinate placement: Placing triangle with A at origin, B on positive x-axis, C in upper half-plane excludes configurations where vertices ordered clockwise. May not matter for distance/parallelism proofs. Can matter for orientation-dependent results (signed area, cross product direction). Always verify.
Overlook degenerate cases: Proof about triangles inscribed in circle may fail when triangle degenerates to diameter plus point on circle. Always check what happens when points coincide, lines become parallel, figures degenerate.
Cite more powerful result than needed: Using law of cosines to prove result that follows from basic angle-chasing obscures proof's logic. May introduce unnecessary assumptions (like cosine function being well-defined). Use simplest sufficient tool.
Miss the converse trap: "If quadrilateral is parallelogram, then its diagonals bisect each other" is true. But its converse is separate theorem requiring separate proof. Don't prove converse when forward direction requested, or vice versa.
Incomplete case analysis: Proof splits into cases (e.g. angle A is acute, right, or obtuse)? All cases must be addressed. Proving acute case and claiming "other cases similar" without verification can hide genuine differences.
construct-geometric-figure - constructions and proofs are complementary: constructions demonstrate existence, proofs establish propertiessolve-trigonometric-problem - trigonometric computations often appear as sub-tasks within geometric proofscreate-skill - follow when packaging new proof technique as reusable skill