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 via Euclidean axiomatic, coordinate, or vector methods w/ rigorous step-by-step logical structure. Direct proof, by contradiction, coordinate, vector, special cases + degenerate configs. Use → geometric statement to prove, verify conjecture, establish lemma, convert intuition to rigorous proof, compare proof method effectiveness.
Independent GitHub snapshot · Automated classification · Original documentation governs
Prove geometric theorem rigorously: pick proof method, construct logical chain of justified steps from hypotheses → conclusion, handle special cases, produce complete proof doc.
Rewrite in standard form w/ explicit Given + Prove clauses.
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.
State conclusion. Exactly what must be proved in "Prove". Distinguish:
Implicit assumptions. Many problems assume Euclidean (parallel postulate), non-degeneracy (points not coincident, lines not concurrent unless stated), positive orientation. Make explicit.
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.
Pick technique best suited to theorem's structure.
Available methods + when to use:
Direct (synthetic): Forward from hypotheses via Euclidean propositions + established theorems.
By contradiction: Assume negation of conclusion, derive contradiction.
Coordinate: Place figure in coordinates + use algebra.
Vector: Geometric relationships as vector ops.
Transformation: Apply geometric transformation (reflection, rotation, translation, dilation) mapping part to part.
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.
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:
→ 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.
Configurations where general argument might fail.
Degenerate cases. Proof holds when:
Boundary cases. Extreme values:
Coordinate proofs → verify coord assignment doesn't lose generality:
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.
Assemble final doc combining all elements.
Header: State theorem in Given/Prove form.
Proof body: Complete chain of justified steps from Step 3.
Special cases: From Step 4 either inline (brief) or remark after main.
Termination: Clear marker:
Review proof for logical completeness:
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.
→ 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.
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 propertiessolve-trigonometric-problem — trig computations often appear as sub-tasks in geometric proofscreate-skill — packaging new proof technique as reusable skill