{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-right-triangle-01",
  "mko_id": "mko-right-triangle",
  "title": {
    "zh-Hant": "直角三角形：用座標判定哪一個內角是直角",
    "en": "Right triangle: deciding in coordinates which interior angle is right"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-right-triangle-01",
    "source": {
      "mko_id": "mko-right-triangle",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "189cad28fbc4231852d6848d7f5209ff5d91a3b2dabd6380708fc678de0a758e",
      "english_sha256": "d07ac2fbb2083e7157e6f68c93440219d29942ce16690399e7ac0757ed7df20d"
    },
    "mode": "definition_walkthrough",
    "audience_prerequisites": [
      "cartesian_product_def",
      "image_preimage_def",
      "square_nonneg",
      "sqrt_def",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在座標平面模型 ℝ² 中，設 A=(1,1)、B=(5,3)、C=(0,3)。證明 A、B、C 構成非退化三角形，且其在 A 的內角等於 90°（即 π/2），所以 ABC 是直角三角形；說明另外兩角不是直角、任何非退化三角形至多一個直角，並指出斜邊與兩股。再以 D=(1,4) 說明三角形 ABD 不是直角三角形。",
      "en": "In the coordinate-plane model ℝ², let A=(1,1), B=(5,3), C=(0,3). Prove that A, B, C form a nondegenerate triangle whose interior angle at A equals 90° (that is, π/2), so ABC is a right triangle; show that the other two angles are not right and that any nondegenerate triangle has at most one right angle, and identify the hypotenuse and the legs. Then, with D=(1,4), show that triangle ABD is not a right triangle."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "模型：點是 ℝ² 的有序實數對，運算逐坐標進行；內積 u·v=u₁v₁+u₂v₂，長度 |u|=√(u₁²+u₂²)；對相異點 P≠Q，直線 L(P,Q)={P+t(Q−P) ∣ t∈ℝ}。",
          "en": "Model: points are ordered pairs of reals in ℝ², with coordinatewise operations; dot product u·v=u₁v₁+u₂v₂ and length |u|=√(u₁²+u₂²); for distinct points P≠Q, the line L(P,Q)={P+t(Q−P) ∣ t∈ℝ}."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "內角量值：對非退化三角形的頂點 X 與另兩頂點 Y、Z，令 u=Y−X、v=Z−X（皆非零），X 處內角的量值是唯一滿足 cos θ=(u·v)/(|u||v|) 的 θ∈[0,π]（弧度）。",
          "en": "Interior angle measure: for a vertex X of a nondegenerate triangle with other vertices Y, Z, let u=Y−X and v=Z−X (both nonzero); the measure of the interior angle at X is the unique θ∈[0,π] (radians) with cos θ=(u·v)/(|u||v|)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "A=(1,1)、B=(5,3)、C=(0,3)、D=(1,4)。",
          "en": "A=(1,1), B=(5,3), C=(0,3), D=(1,4)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G2",
          "mko_right_triangle_def"
        ],
        "before_tex": "\\triangle ABC\\ \\text{is a right triangle}",
        "after_tex": "\\triangle ABC\\ \\text{nondegenerate and some interior angle}=90^\\circ=\\tfrac{\\pi}{2}",
        "relation": "definition",
        "rule": "mko_right_triangle_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，直角三角形是一個內角等於 90°（或弧度 π/2）的三角形，而且討論對象是平面中的非退化三角形（假設 a1）。所以要判定 ABC 是直角三角形，必須做兩件事：先確認 A、B、C 構成非退化三角形，再找出一個等於 π/2 的內角。內角量值用 G2 的定義。",
          "en": "By the MKO, a right triangle is a triangle with an interior angle of 90° (π/2 in radians), and the object discussed is a nondegenerate triangle in the plane (assumption a1). So deciding that ABC is a right triangle takes two tasks: first confirm that A, B, C form a nondegenerate triangle, then find an interior angle equal to π/2. Interior angles are measured by the definition in G2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_right_triangle_def",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義，尚未推出任何事實；順序很重要：沒有非退化三角形，就沒有內角可談。",
            "en": "This step only states the definition and derives nothing yet; the order matters: without a nondegenerate triangle there are no interior angles to speak of."
          }
        },
        "leaf_basis": "mko_right_triangle_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "deg_rad"
        ],
        "before_tex": "90^\\circ",
        "after_tex": "\\tfrac{\\pi}{2}\\ \\text{rad}",
        "relation": "equivalent",
        "rule": "deg_rad",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依換算 θ 度=θ·π/180 弧度：90·π/180=(90/180)·π=π/2。G2 的量值以弧度表示，所以之後的目標是 θ=π/2。",
          "en": "By the conversion θ degrees = θ·π/180 radians: 90·π/180=(90/180)·π=π/2. The measures in G2 are in radians, so the target from now on is θ=π/2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "deg_rad",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：(π/2)·(180/π)=90，回到 90°；兩種寫法表示同一角度。",
            "en": "Backward: (π/2)·(180/π)=90, giving back 90°; both forms denote the same angle."
          }
        },
        "leaf_basis": "deg_rad + frac_rules + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "G3",
          "cartesian_product_def"
        ],
        "before_tex": "A=(1,1),\\ B=(5,3),\\ C=(0,3)",
        "after_tex": "A\\neq B,\\ B\\neq C,\\ C\\neq A",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第一坐標 1≠5、5≠0、0≠1，所以三點兩兩相異。特別地，u=B−A 與 v=C−A 都不是零向量，這是 G2 與之後引理所需的條件。",
          "en": "The first coordinates give 1≠5, 5≠0 and 0≠1, so the points are pairwise distinct. In particular u=B−A and v=C−A are nonzero vectors, which G2 and the lemma below require."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "一個坐標不同即可推出兩點不等；三對都要檢查。",
            "en": "One differing coordinate shows two points differ; all three pairs are checked."
          }
        },
        "leaf_basis": "cartesian_product_def + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "G3",
          "S03"
        ],
        "before_tex": "C\\in L(A,B)",
        "after_tex": "\\exists t\\in\\mathbb{R}:\\ t=-\\tfrac{1}{4}\\ \\land\\ t=1",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [
          "4\\neq 0",
          "2\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "B−A=(4,2)，L(A,B)={(1+4t, 1+2t) ∣ t∈ℝ}。C∈L(A,B) ⇔ 有 t 使 (0,3)=(1+4t, 1+2t) ⇔ 0=1+4t 且 3=1+2t。第一式兩邊減 1 再除以 4（4≠0）得 t=−1/4；第二式兩邊減 1 再除以 2（2≠0）得 t=1。每個變形都是等價。",
          "en": "B−A=(4,2), so L(A,B)={(1+4t, 1+2t) ∣ t∈ℝ}. C∈L(A,B) ⇔ some t gives (0,3)=(1+4t, 1+2t) ⇔ 0=1+4t and 3=1+2t. Subtracting 1 and dividing by 4 (4≠0) turns the first into t=−1/4; subtracting 1 and dividing by 2 (2≠0) turns the second into t=1. Each move is an equivalence."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "image_preimage_def",
            "cartesian_product_def",
            "eq_congruence",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "代回：1+4·(−1/4)=0、1+2·1=3；每個變形可逆（除數非零）。",
            "en": "Substituting back: 1+4·(−1/4)=0 and 1+2·1=3; each move is reversible (the divisors are nonzero)."
          }
        },
        "leaf_basis": "image_preimage_def + cartesian_product_def + eq_congruence + mul_inverse_nonzero"
      },
      {
        "id": "S05",
        "uses": [
          "S04"
        ],
        "before_tex": "\\exists t:\\ t=-\\tfrac{1}{4}\\ \\land\\ t=1",
        "after_tex": "C\\notin L(A,B)",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若同一個 t 同時等於 −1/4 與 1，依等號的對稱與傳遞得 −1/4=1，但 −1/4≠1。所以沒有這樣的 t，依 S04，C∉L(A,B)。",
          "en": "If one t equalled both −1/4 and 1, symmetry and transitivity of equality would give −1/4=1, but −1/4≠1. So no such t exists, and by S04, C∉L(A,B)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "contradiction",
            "eq_refl_symm_trans"
          ],
          "explanation": {
            "zh-Hant": "每個坐標方程單獨都有解；矛盾來自兩式要用同一個 t。",
            "en": "Each coordinate equation alone has a solution; the contradiction comes from both needing the same t."
          }
        },
        "leaf_basis": "contradiction + eq_refl_symm_trans + numeric_evaluation"
      },
      {
        "id": "S06",
        "uses": [
          "S03",
          "S05",
          "line_unique",
          "mko_triangle_def"
        ],
        "before_tex": "A\\neq B,\\ C\\notin L(A,B)",
        "after_tex": "\\triangle ABC\\ \\text{nondegenerate}",
        "relation": "implies",
        "rule": "line_unique",
        "conditions": [
          "A\\neq B"
        ],
        "explanation": {
          "zh-Hant": "若某條直線同時包含 A、B、C，因 A≠B，依 line_unique 它就是 L(A,B)，於是 C∈L(A,B)，與 S05 矛盾。所以 A、B、C 不共線；加上 S03 的相異，依三角形的 MKO，ABC 是非退化三角形，MKO 假設 a1 成立。",
          "en": "If some line contained A, B and C, then since A≠B, by line_unique that line is L(A,B), so C∈L(A,B), contradicting S05. So A, B, C are noncollinear; together with distinctness from S03, the triangle MKO makes ABC a nondegenerate triangle, and the MKO's assumption a1 holds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S05",
            "line_unique",
            "collinear_def",
            "mko_triangle_def",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "唯一性是關鍵：只有 C∉L(A,B)，若不知道通過 A、B 的直線只有一條，就不能排除其他直線。",
            "en": "Uniqueness is the key: C∉L(A,B) alone does not rule out other lines unless the line through A and B is unique."
          }
        },
        "leaf_basis": "line_unique + collinear_def + contradiction + mko_triangle_def"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "G3",
          "S03"
        ],
        "before_tex": "u=B-A,\\ v=C-A",
        "after_tex": "u=(4,2),\\ v=(-1,2),\\ u\\cdot v=0",
        "relation": "equivalent",
        "rule": "dot_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐坐標：u=(5−1, 3−1)=(4,2)，v=(0−1, 3−1)=(−1,2)。內積 u·v=4·(−1)+2·2=−4+4=0。這是精確的整數計算，沒有誤差。",
          "en": "Coordinatewise: u=(5−1, 3−1)=(4,2) and v=(0−1, 3−1)=(−1,2). The dot product is u·v=4·(−1)+2·2=−4+4=0. This is an exact integer computation with no error."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "dot_def",
            "coord_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：A+u=(1+4,1+2)=(5,3)=B、A+v=(1−1,1+2)=(0,3)=C；內積可獨立重算（N5）。",
            "en": "Backward: A+u=(1+4,1+2)=(5,3)=B and A+v=(1−1,1+2)=(0,3)=C; the dot product can be recomputed independently (N5)."
          }
        },
        "leaf_basis": "dot_def + coord_ops + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S06",
          "S07",
          "L_right_iff_dot"
        ],
        "before_tex": "u\\cdot v=0,\\ u\\neq 0,\\ v\\neq 0",
        "after_tex": "\\theta_A=\\tfrac{\\pi}{2}",
        "relation": "equivalent",
        "rule": "L_right_iff_dot",
        "conditions": [
          "u\\neq 0",
          "v\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "ABC 非退化（S06），所以 G2 給出 A 處內角 θ_A∈[0,π]，cos θ_A=(u·v)/(|u||v|)，且 u、v 非零（S03）。依引理 L_right_iff_dot，u·v=0 ⇔ θ_A=π/2，所以 θ_A=π/2。引理在 S08.1、S08.2 證明。",
          "en": "ABC is nondegenerate (S06), so G2 gives the interior angle θ_A∈[0,π] at A with cos θ_A=(u·v)/(|u||v|), and u, v are nonzero (S03). By Lemma L_right_iff_dot, u·v=0 ⇔ θ_A=π/2, so θ_A=π/2. The lemma is proved in S08.1 and S08.2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_right_iff_dot",
            "S03",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "引理是雙條件，兩個方向都在子步證明；u、v 非零的條件不可省略。",
            "en": "The lemma is a biconditional with both directions proved in the sub-steps; the condition that u and v are nonzero cannot be dropped."
          }
        },
        "leaf_basis": "L_right_iff_dot + angle_cos_def"
      },
      {
        "id": "S08.1",
        "uses": [
          "dot_def",
          "square_nonneg"
        ],
        "before_tex": "\\cos\\theta=\\frac{u\\cdot v}{|u|\\,|v|},\\ u\\neq 0,\\ v\\neq 0",
        "after_tex": "\\cos\\theta=0\\iff u\\cdot v=0",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "|u|\\,|v|\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "u≠0 表示 u₁、u₂ 不全為 0；平方非負，且非零數的平方大於 0，所以 u₁²+u₂²>0，|u|=√(u₁²+u₂²)>0。同理 |v|>0，故 |u||v|>0，特別是 ≠0。於是 cos θ=0 時兩邊乘 |u||v| 得 u·v=0；反之 u·v=0 時 cos θ=0/(|u||v|)=0。",
          "en": "u≠0 means u₁ and u₂ are not both 0; squares are nonnegative and the square of a nonzero number is positive, so u₁²+u₂²>0 and |u|=√(u₁²+u₂²)>0. Likewise |v|>0, so |u||v|>0, in particular ≠0. If cos θ=0, multiplying by |u||v| gives u·v=0; conversely, if u·v=0 then cos θ=0/(|u||v|)=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "square_nonneg",
            "sqrt_def"
          ],
          "explanation": {
            "zh-Hant": "乘以 |u||v| 與除以 |u||v| 互為反運算，前提 |u||v|≠0；若 u=0，分母為 0，cos θ 的式子沒有定義。",
            "en": "Multiplying and dividing by |u||v| undo each other, given |u||v|≠0; if u=0 the denominator is 0 and the formula for cos θ is undefined."
          }
        },
        "leaf_basis": "square_nonneg + sqrt_def + order_basics + mul_inverse_nonzero + mul_zero"
      },
      {
        "id": "S08.2",
        "uses": [
          "cos_facts",
          "S08.1"
        ],
        "before_tex": "\\theta\\in[0,\\pi]",
        "after_tex": "\\theta=\\tfrac{\\pi}{2}\\iff\\cos\\theta=0\\iff u\\cdot v=0",
        "relation": "equivalent",
        "rule": "cos_facts",
        "conditions": [
          "\\theta\\in[0,\\pi]"
        ],
        "explanation": {
          "zh-Hant": "⇒：cos(π/2)=0。⇐：cos 在 [0,π] 上嚴格遞減，所以在那裡單射；若 θ∈[0,π] 且 cos θ=0=cos(π/2)，則 θ=π/2。再與 S08.1 串接得 θ=π/2 ⇔ u·v=0。S08.1 與 S08.2 合起來證得引理 L_right_iff_dot。",
          "en": "⇒: cos(π/2)=0. ⇐: cos is strictly decreasing on [0,π], hence injective there; if θ∈[0,π] and cos θ=0=cos(π/2), then θ=π/2. Chaining with S08.1 gives θ=π/2 ⇔ u·v=0. S08.1 and S08.2 together prove Lemma L_right_iff_dot."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cos_facts",
            "logic_chain",
            "S08.1"
          ],
          "explanation": {
            "zh-Hant": "⇐ 方向需要 θ 限制在 [0,π]；若允許 θ=3π/2，cos θ=0 也成立，所以 G2 的取值範圍不可省略。",
            "en": "The ⇐ direction needs θ restricted to [0,π]; if θ=3π/2 were allowed, cos θ=0 would hold too, so the range in G2 cannot be dropped."
          }
        },
        "leaf_basis": "cos_facts + logic_chain + injective_surjective_def"
      },
      {
        "id": "S09",
        "uses": [
          "S02",
          "S06",
          "S08",
          "mko_right_triangle_def"
        ],
        "before_tex": "\\triangle ABC\\ \\text{nondegenerate},\\ \\theta_A=\\tfrac{\\pi}{2}",
        "after_tex": "\\triangle ABC\\ \\text{is a right triangle with the right angle at}\\ A",
        "relation": "implies",
        "rule": "mko_right_triangle_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非退化三角形 ABC 有一個內角 θ_A=π/2=90°（S02、S08），符合 MKO 的定義，所以 ABC 是直角三角形，直角在 A。",
          "en": "The nondegenerate triangle ABC has an interior angle θ_A=π/2=90° (S02, S08), which meets the MKO's definition, so ABC is a right triangle with the right angle at A."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S06",
            "S08",
            "mko_right_triangle_def",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "定義只要求存在一個直角；兩個要求（非退化、有一角為 π/2）都已證。",
            "en": "The definition asks only for one right angle to exist; both requirements (nondegenerate, one angle equal to π/2) are proved."
          }
        },
        "leaf_basis": "mko_right_triangle_def + exists_intro + and_intro_elim"
      },
      {
        "id": "S10",
        "uses": [
          "S03",
          "S06",
          "G3",
          "L_right_iff_dot"
        ],
        "before_tex": "(A-B)\\cdot(C-B)=(-4,-2)\\cdot(-5,0)",
        "after_tex": "20\\neq 0,\\ \\theta_B\\neq\\tfrac{\\pi}{2}",
        "relation": "equivalent",
        "rule": "L_right_iff_dot",
        "conditions": [
          "A-B\\neq 0",
          "C-B\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "在 B：A−B=(1−5, 1−3)=(−4,−2)，C−B=(0−5, 3−3)=(−5,0)，皆非零；內積 (−4)(−5)+(−2)·0=20≠0。引理是雙條件，兩邊同時否定仍等價：內積 ≠0 ⇔ θ_B≠π/2。",
          "en": "At B: A−B=(1−5, 1−3)=(−4,−2) and C−B=(0−5, 3−3)=(−5,0), both nonzero; the dot product is (−4)(−5)+(−2)·0=20≠0. The lemma is a biconditional, so negating both sides keeps an equivalence: dot product ≠0 ⇔ θ_B≠π/2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_right_iff_dot",
            "dot_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "P⇔Q 與 ¬P⇔¬Q 等價；計算可重算（N6）。",
            "en": "P⇔Q is equivalent to ¬P⇔¬Q; the computation can be redone (N6)."
          }
        },
        "leaf_basis": "L_right_iff_dot + dot_def + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S03",
          "S06",
          "G3",
          "L_right_iff_dot"
        ],
        "before_tex": "(A-C)\\cdot(B-C)=(1,-2)\\cdot(5,0)",
        "after_tex": "5\\neq 0,\\ \\theta_C\\neq\\tfrac{\\pi}{2}",
        "relation": "equivalent",
        "rule": "L_right_iff_dot",
        "conditions": [
          "A-C\\neq 0",
          "B-C\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "在 C：A−C=(1−0, 1−3)=(1,−2)，B−C=(5−0, 3−3)=(5,0)，皆非零；內積 1·5+(−2)·0=5≠0，所以 θ_C≠π/2。",
          "en": "At C: A−C=(1−0, 1−3)=(1,−2) and B−C=(5−0, 3−3)=(5,0), both nonzero; the dot product is 1·5+(−2)·0=5≠0, so θ_C≠π/2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_right_iff_dot",
            "dot_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "與 S10 相同，引理的雙條件兩邊同時否定（N7）。",
            "en": "As in S10, both sides of the lemma's biconditional are negated (N7)."
          }
        },
        "leaf_basis": "L_right_iff_dot + dot_def + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "thm_angle_sum",
          "angle_positive"
        ],
        "before_tex": "\\theta_X=\\theta_Y=\\tfrac{\\pi}{2}\\ \\text{at two vertices}\\ X\\neq Y",
        "after_tex": "\\text{contradiction; a nondegenerate triangle has at most one right angle}",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對任意非退化三角形：若兩個內角都等於 π/2，依內角和定理，第三個角等於 π−π/2−π/2=0；但非退化三角形的每個內角都大於 0（angle_positive），矛盾。所以至多一個直角。這個論證不依賴具體座標，對每個非退化三角形成立；它與 S10、S11 的直接計算一致。",
          "en": "For any nondegenerate triangle: if two interior angles both equal π/2, the angle-sum theorem makes the third π−π/2−π/2=0; but every interior angle of a nondegenerate triangle is greater than 0 (angle_positive), a contradiction. So there is at most one right angle. The argument uses no particular coordinates and holds for every nondegenerate triangle; it agrees with the direct computations in S10 and S11."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "thm_angle_sum",
            "angle_positive",
            "contradiction",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要內角和定理（外部引用，citation_verified）與每角為正；只靠一個例子的計算，不能推出「所有非退化三角形至多一個直角」。",
            "en": "It needs the angle-sum theorem (cited from outside, citation_verified) and positivity of each angle; computations for one example could not establish that every nondegenerate triangle has at most one right angle."
          }
        },
        "leaf_basis": "thm_angle_sum + angle_positive + contradiction + forall_intro"
      },
      {
        "id": "S13",
        "uses": [
          "S09",
          "dot_def"
        ],
        "before_tex": "\\theta_A=\\tfrac{\\pi}{2}",
        "after_tex": "\\text{hypotenuse}\\ BC,\\ \\text{legs}\\ AB,\\ AC",
        "relation": "definition",
        "rule": "def_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "直角所對的邊稱為斜邊，另外兩邊稱為股（MKO 說明）。A 的對邊是 BC，所以 BC 是斜邊，AB、AC 是股。核對長度平方：|AB|²=4²+2²=20，|AC|²=(−1)²+2²=5，|BC|²=(−5)²+0²=25；本例中斜邊最長（25>20>5）。這只是此例的核對，不是「斜邊最長」的一般證明。",
          "en": "The side opposite the right angle is called the hypotenuse and the other two sides are the legs (MKO explanation). The side opposite A is BC, so BC is the hypotenuse and AB, AC are the legs. Checking squared lengths: |AB|²=4²+2²=20, |AC|²=(−1)²+2²=5, |BC|²=(−5)²+0²=25; in this example the hypotenuse is longest (25>20>5). This is a check for this example only, not a general proof that the hypotenuse is longest."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S09",
            "def_intro",
            "dot_def"
          ],
          "explanation": {
            "zh-Hant": "命名只依賴直角的位置（S09）；長度平方只作核對，不參與推論。",
            "en": "The naming depends only on where the right angle is (S09); the squared lengths are a check and play no part in the inference."
          }
        },
        "leaf_basis": "def_intro + dot_def + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "G1",
          "G3",
          "line_unique"
        ],
        "before_tex": "A,\\ B,\\ D=(1,4)",
        "after_tex": "\\triangle ABD\\ \\text{nondegenerate}",
        "relation": "implies",
        "rule": "line_unique",
        "conditions": [
          "A\\neq B"
        ],
        "explanation": {
          "zh-Hant": "D 與 A、B 相異（D 的第二坐標 4 與 1、3 不同）。D∈L(A,B) ⇔ 有 t 使 1=1+4t 且 4=1+2t ⇔ t=0 且 t=3/2，不可能；所以 D∉L(A,B)。與 S06 同理，依直線唯一性 A、B、D 不共線，ABD 是非退化三角形。",
          "en": "D is distinct from A and B (its second coordinate 4 differs from 1 and 3). D∈L(A,B) ⇔ some t gives 1=1+4t and 4=1+2t ⇔ t=0 and t=3/2, which is impossible; so D∉L(A,B). As in S06, uniqueness of the line makes A, B, D noncollinear, so ABD is a nondegenerate triangle."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "line_unique",
            "contradiction",
            "mko_triangle_def"
          ],
          "explanation": {
            "zh-Hant": "與 S04 到 S06 相同的三個檢查：相異、不在 L(A,B) 上、直線唯一。",
            "en": "The same three checks as S04 to S06: distinctness, not on L(A,B), and uniqueness of the line."
          }
        },
        "leaf_basis": "line_unique + image_preimage_def + contradiction + mko_triangle_def"
      },
      {
        "id": "S15",
        "uses": [
          "S14",
          "L_right_iff_dot",
          "G3"
        ],
        "before_tex": "(B-A)\\cdot(D-A),\\ (A-B)\\cdot(D-B),\\ (A-D)\\cdot(B-D)",
        "after_tex": "6,\\ 14,\\ 3\\neq 0;\\ \\triangle ABD\\ \\text{not right}",
        "relation": "implies",
        "rule": "L_right_iff_dot",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 A：(4,2)·(0,3)=0+6=6；在 B：(−4,−2)·(−4,1)=16−2=14；在 D：(0,−3)·(4,−1)=0+3=3。三個內積都非零，依引理三個內角都不是 π/2，所以 ABD 沒有直角，依 MKO 不是直角三角形。AD 是鉛直的，但只有一邊鉛直不能決定直角。",
          "en": "At A: (4,2)·(0,3)=0+6=6; at B: (−4,−2)·(−4,1)=16−2=14; at D: (0,−3)·(4,−1)=0+3=3. All three dot products are nonzero, so by the lemma no interior angle is π/2; ABD has no right angle and by the MKO is not a right triangle. AD is vertical, but one vertical side alone does not decide a right angle."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S14",
            "L_right_iff_dot",
            "dot_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "要證「不是直角三角形」，三個角都必須排除；只排除一個不夠。",
            "en": "To show a triangle is not right, all three angles must be excluded; excluding one is not enough."
          }
        },
        "leaf_basis": "L_right_iff_dot + dot_def + numeric_evaluation + de_morgan_logic"
      },
      {
        "id": "S16",
        "uses": [
          "S09",
          "S10",
          "S11",
          "S12",
          "S13",
          "S15"
        ],
        "before_tex": "\\text{S09, S12, S13, S15}",
        "after_tex": "\\triangle ABC\\ \\text{right at}\\ A;\\ \\triangle ABD\\ \\text{not right}",
        "relation": "implies",
        "rule": "mko_right_triangle_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "ABC 是直角三角形，直角在 A，另兩角不是直角，斜邊 BC、股 AB 與 AC；ABD 是非退化三角形但沒有直角。",
          "en": "ABC is a right triangle with the right angle at A; the other two angles are not right; the hypotenuse is BC and the legs are AB and AC. ABD is a nondegenerate triangle with no right angle."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S06",
            "S08",
            "S14",
            "S15",
            "mko_right_triangle_def"
          ],
          "explanation": {
            "zh-Hant": "目標：ABC 是直角三角形而 ABD 不是。足夠條件 H：ABC 非退化且 θ_A=π/2（S06、S08），ABD 非退化且三個角都不是 π/2（S14、S15）；H 推出目標由 MKO 的定義。H 由前面各步正向證成，沒有把目標當作前提。",
            "en": "Goal: ABC is a right triangle and ABD is not. Sufficient condition H: ABC is nondegenerate with θ_A=π/2 (S06, S08), and ABD is nondegenerate with no angle equal to π/2 (S14, S15); H gives the goal by the MKO's definition. H is proved forwards in the earlier steps; the goal is never used as a premise."
          }
        },
        "leaf_basis": "mko_right_triangle_def + and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "A=(1,1)、B=(5,3)、C=(0,3) 構成非退化三角形；u=B−A=(4,2)、v=C−A=(−1,2) 的內積為 0，所以 θ_A=π/2=90°，ABC 是直角三角形，直角在 A，斜邊是 BC，股是 AB 與 AC。B、C 處的內積為 20、5，不是直角；任何非退化三角形至多一個直角。三角形 ABD（D=(1,4)）三個內積為 6、14、3，沒有直角，不是直角三角形。",
      "en": "A=(1,1), B=(5,3), C=(0,3) form a nondegenerate triangle; u=B−A=(4,2) and v=C−A=(−1,2) have dot product 0, so θ_A=π/2=90° and ABC is a right triangle with the right angle at A, hypotenuse BC and legs AB, AC. The dot products at B and C are 20 and 5, so those angles are not right; any nondegenerate triangle has at most one right angle. Triangle ABD (D=(1,4)) has dot products 6, 14 and 3, so it has no right angle and is not a right triangle."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "E=(1,1) 時 (B−A)·(E−A)=0，所以 ABE 是直角三角形，直角在 A",
          "en": "With E=(1,1), (B−A)·(E−A)=0, so ABE is a right triangle with the right angle at A"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "E=A，三點不相異，E−A 是零向量；沒有非退化三角形，也沒有角（MKO 假設 a1、G2）。",
          "en": "E=A, so the points are not distinct and E−A is the zero vector; there is no nondegenerate triangle and no angle (MKO assumption a1, G2)."
        }
      },
      {
        "claim": {
          "zh-Hant": "三角形 ABD 的邊 AD 是鉛直的，所以 ABD 有直角",
          "en": "Side AD of triangle ABD is vertical, so ABD has a right angle"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "三個頂點的內積為 6、14、3，都非零（S15），沒有任何內角等於 π/2。",
          "en": "The dot products at the three vertices are 6, 14 and 3, all nonzero (S15), so no interior angle equals π/2."
        }
      },
      {
        "claim": {
          "zh-Hant": "某些非退化三角形有兩個直角",
          "en": "Some nondegenerate triangles have two right angles"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "若有兩個直角，第三角即為 π−π/2−π/2=0，與每個內角大於 0 矛盾（S12）。",
          "en": "Then the third angle would be π−π/2−π/2=0, contradicting that every interior angle is positive (S12)."
        }
      },
      {
        "claim": {
          "zh-Hant": "ABC 的直角在 A，所以斜邊是 AB",
          "en": "The right angle of ABC is at A, so the hypotenuse is AB"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "斜邊是直角所對的邊，即 BC；AB 以 A 為端點，是股。本例 |BC|²=25 也大於 |AB|²=20。",
          "en": "The hypotenuse is the side opposite the right angle, namely BC; AB has A as an endpoint and is a leg. In this example |BC|²=25 is also larger than |AB|²=20."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "id": "image_preimage_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。",
          "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse."
        }
      },
      {
        "id": "square_nonneg",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。",
          "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0."
        }
      },
      {
        "id": "sqrt_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "主平方根：對 a≥0，√a 是唯一滿足 b≥0 且 b²=a 的實數 b。負數在實數中沒有平方根。",
          "en": "Principal square root: for a≥0, √a is the unique real b with b≥0 and b²=a. Negative numbers have no real square root."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "mul_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。",
          "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0."
        }
      },
      {
        "id": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "id": "injective_surjective_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "單射：f(a)=f(a′) ⇒ a=a′。滿射：每個 b∈B 都有 a∈A 使 f(a)=b。雙射：既單又滿，此時有唯一反函數 f⁻¹ 使 f⁻¹∘f=id_A、f∘f⁻¹=id_B。",
          "en": "Injective: f(a)=f(a′) ⇒ a=a′. Surjective: every b∈B has some a∈A with f(a)=b. Bijective: both; then there is a unique inverse f⁻¹ with f⁻¹∘f=id_A and f∘f⁻¹=id_B."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "def_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "id": "de_morgan_logic",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "邏輯的德摩根律：¬(P∧Q) ⇔ ¬P∨¬Q；¬(P∨Q) ⇔ ¬P∧¬Q。",
          "en": "De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q."
        }
      },
      {
        "id": "mko_right_triangle_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-right-triangle",
          "sha256": "189cad28fbc4231852d6848d7f5209ff5d91a3b2dabd6380708fc678de0a758e",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "若三角形的一個內角等於九十度，或以弧度表示為 π/2，則此三角形稱為直角三角形。",
          "en": "A triangle is called a right triangle if one interior angle equals ninety degrees, equivalently π/2 radians."
        }
      },
      {
        "id": "mko_triangle_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-triangle",
          "sha256": "1556dabd42bf5d132f41296684eb2b037900a511e659c48627bee6d0c4dcd461",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給定平面中三個相異且不共線的點 A、B、C，線段 AB、BC、CA 的聯合界定非退化三角形 ABC。",
          "en": "Given three distinct noncollinear points A, B, and C in the plane, the union of segments AB, BC, and CA determines the nondegenerate triangle ABC."
        }
      },
      {
        "id": "deg_rad",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "度數與弧度：θ 度等於 θ·π/180 弧度；90° 與 π/2 表示同一角度（MKO 假設 a2）。",
          "en": "Degrees and radians: θ degrees equals θ·π/180 radians; 90° and π/2 denote the same angle (MKO assumption a2)."
        }
      },
      {
        "id": "dot_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "內積與長度：u·v=u₁v₁+u₂v₂，|u|=√(u₁²+u₂²)。",
          "en": "Dot product and length: u·v=u₁v₁+u₂v₂ and |u|=√(u₁²+u₂²)."
        }
      },
      {
        "id": "coord_ops",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逐坐標運算：(a,b)+(c,d)=(a+c,b+d)，(a,b)−(c,d)=(a−c,b−d)，t(a,b)=(ta,tb)。",
          "en": "Coordinatewise operations: (a,b)+(c,d)=(a+c,b+d), (a,b)−(c,d)=(a−c,b−d), t(a,b)=(ta,tb)."
        }
      },
      {
        "id": "angle_cos_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "座標模型中的內角量值：頂點 X、u=Y−X、v=Z−X 非零時，X 處內角是唯一滿足 cos θ=(u·v)/(|u||v|) 的 θ∈[0,π]。",
          "en": "Interior angle measure in the coordinate model: at a vertex X with nonzero u=Y−X and v=Z−X, the interior angle is the unique θ∈[0,π] with cos θ=(u·v)/(|u||v|)."
        }
      },
      {
        "id": "cos_facts",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "餘弦函數在 [0,π] 上嚴格遞減（因此在 [0,π] 上單射），且 cos(π/2)=0。",
          "en": "The cosine function is strictly decreasing on [0,π] (hence injective there), and cos(π/2)=0."
        }
      },
      {
        "id": "line_unique",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "在 ℝ² 中，通過相異兩點 X、Y 的直線恰有一條，就是 L(X,Y)。",
          "en": "In ℝ², exactly one line passes through two distinct points X and Y, namely L(X,Y)."
        }
      },
      {
        "id": "collinear_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "共線：若存在一條直線同時包含若干點，這些點稱為共線。",
          "en": "Collinear: points are collinear if some line contains all of them."
        }
      },
      {
        "id": "thm_angle_sum",
        "kind": "cited",
        "source": {
          "reference": "Euclid, Elements, Book I, Proposition 32",
          "original_reference": "Euclid, Elements, Book I, Proposition 32",
          "url": "https://www.euclids-elements.org/elements/books/bookI/propositions/propI32/",
          "locator": "Proposition 32, statement and proof",
          "checked_date": "2026-10-19",
          "verification": "original_verified",
          "verification_note": "Euclidean nondegenerate triangles; two right angles."
        },
        "needs_citation": false,
        "statement": {
          "zh-Hant": "歐幾里得平面中，非退化三角形的三個內角之和等於兩個直角，即 π（180°）。",
          "en": "In the Euclidean plane, the three interior angles of a nondegenerate triangle together equal two right angles, that is, π (180°)."
        }
      },
      {
        "id": "angle_positive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非退化三角形的每個內角量值都嚴格介於 0 與 π 之間。",
          "en": "Each interior angle of a nondegenerate triangle has measure strictly between 0 and π."
        }
      },
      {
        "id": "L_right_iff_dot",
        "kind": "lemma",
        "proved_in": [
          "S08.1",
          "S08.2"
        ],
        "statement": {
          "zh-Hant": "對非零向量 u、v 與 θ∈[0,π]，若 cos θ=(u·v)/(|u||v|)，則 θ=π/2 ⇔ u·v=0。",
          "en": "For nonzero vectors u, v and θ∈[0,π] with cos θ=(u·v)/(|u||v|): θ=π/2 ⇔ u·v=0."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "90/180",
        "rhs": "1/2"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "(0-1)/4",
        "rhs": "-1/4"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "(3-1)/2",
        "rhs": "1"
      },
      {
        "id": "N4",
        "step": "S05",
        "lhs": "-1/4",
        "relation": "!=",
        "rhs": "1"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "4*(-1)+2*2",
        "rhs": "0"
      },
      {
        "id": "N6",
        "step": "S10",
        "lhs": "(-4)*(-5)+(-2)*0",
        "rhs": "20"
      },
      {
        "id": "N7",
        "step": "S11",
        "lhs": "1*5+(-2)*0",
        "rhs": "5"
      },
      {
        "id": "N8",
        "step": "S13",
        "lhs": "4**2+2**2",
        "rhs": "20"
      },
      {
        "id": "N9",
        "step": "S13",
        "lhs": "(-1)**2+2**2",
        "rhs": "5"
      },
      {
        "id": "N10",
        "step": "S13",
        "lhs": "(-5)**2+0**2",
        "rhs": "25"
      },
      {
        "id": "N11",
        "step": "S12",
        "lhs": "180-90-90",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S14",
        "lhs": "(1-1)/4",
        "rhs": "0"
      },
      {
        "id": "N13",
        "step": "S14",
        "lhs": "(4-1)/2",
        "rhs": "3/2"
      },
      {
        "id": "N14",
        "step": "S15",
        "lhs": "4*0+2*3",
        "rhs": "6"
      },
      {
        "id": "N15",
        "step": "S15",
        "lhs": "(-4)*(-4)+(-2)*1",
        "rhs": "14"
      },
      {
        "id": "N16",
        "step": "S15",
        "lhs": "0*4+(-3)*(-1)",
        "rhs": "3"
      },
      {
        "id": "N17",
        "lhs": "20+5",
        "rhs": "25"
      },
      {
        "id": "N18",
        "lhs": "4*0+2*0",
        "rhs": "0"
      },
      {
        "id": "N19",
        "step": "S07",
        "lhs": "1+4",
        "rhs": "5"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 ℝ² 座標模型中工作，內角量值以 G2 的餘弦式定義，取值於 [0,π]；直接使用 MKO 的定義（一個內角等於 90°=π/2），並先確認非退化三角形（MKO 假設 a1）。內角和定理以外部文獻引用（citation_verified），餘弦函數與直線唯一性作為前提。不證明畢氏定理，也不證明「斜邊最長」的一般敘述；不改動 canonical MKO 或英文 edition。",
      "en": "Works in the coordinate model of ℝ², with interior angle measures defined by the cosine formula of G2 and taking values in [0,π]; it uses the MKO's definition directly (one interior angle equals 90°=π/2) and first confirms a nondegenerate triangle (MKO assumption a1). The angle-sum theorem is cited from outside (citation_verified); facts about cosine and uniqueness of lines are prerequisites. It proves neither the Pythagorean theorem nor the general claim that the hypotenuse is longest, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "u\\cdot v",
        "內積 u₁v₁+u₂v₂",
        "dot product u₁v₁+u₂v₂"
      ],
      [
        "|u|",
        "向量長度 √(u₁²+u₂²)",
        "length of a vector, √(u₁²+u₂²)"
      ],
      [
        "\\theta_A,\\theta_B,\\theta_C",
        "三角形在 A、B、C 的內角量值（弧度）",
        "the interior angle measures at A, B, C (radians)"
      ],
      [
        "L(P,Q)",
        "通過相異點 P、Q 的直線",
        "the line through distinct points P and Q"
      ],
      [
        "90^\\circ,\\ \\tfrac{\\pi}{2}",
        "同一個角的度數與弧度寫法",
        "the same angle in degrees and in radians"
      ]
    ],
    "md_error_controls": [
      {
        "title": "先確認非退化三角形 / First confirm a nondegenerate triangle",
        "zh": "MKO 指出的誤解是只看到兩條邊互相垂直，卻沒有確認它們與第三邊構成非退化三角形。例如 E=(1,1)=A 時 (B−A)·(E−A)=(4,2)·(0,0)=0（N18），內積為 0 只是因為 E−A 是零向量，A、B、E 根本不是三個相異點，沒有角可談。又如線段 [(0,0),(2,0)] 與 [(5,1),(5,3)] 方向互相垂直，但它們沒有共同端點，不是同一個三角形的兩邊。所以 S03 到 S06 必須在算角之前完成。",
        "en": "The misconception identified in the MKO is seeing two sides perpendicular without confirming that they form a nondegenerate triangle with the third side. For example, with E=(1,1)=A we get (B−A)·(E−A)=(4,2)·(0,0)=0 (N18), but the dot product is 0 only because E−A is the zero vector; A, B, E are not three distinct points and there is no angle at all. Likewise the segments [(0,0),(2,0)] and [(5,1),(5,3)] have perpendicular directions but share no endpoint, so they are not two sides of one triangle. That is why S03 to S06 must come before any angle is computed."
      },
      {
        "title": "度數與弧度不可混用 / Do not mix degrees and radians",
        "zh": "90° 與 π/2 弧度是同一角度（S02），但數字 90 與 π/2 不相等。若把 θ=90 代入以弧度為單位的 G2，它表示 90 弧度，不是直角：因為 π<4，π/2<2<90。寫角度時必須標明單位。",
        "en": "90° and π/2 radians are the same angle (S02), but the numbers 90 and π/2 are not equal. Putting θ=90 into G2, whose unit is the radian, means 90 radians, which is not a right angle: since π<4, π/2<2<90. Angles must be written with their unit."
      },
      {
        "title": "程式的容差檢查只是具體輸入檢查 / A program's tolerance check is a check of concrete input",
        "zh": "MKO 說明程式可以檢查三個輸入角是否在容差內加總為 180°、其中是否有 90°，但這只是對具體輸入的檢查。本稿用整數內積做精確判定（0、20、5），不需容差；而「至多一個直角」是 S12 的一般論證，不能由任何有限個例子的計算取代。",
        "en": "The MKO notes that a program can check whether three input angles sum to 180° within a tolerance and whether one is 90°, but that is only a check of concrete input. This guide decides exactly with integer dot products (0, 20, 5), so no tolerance is needed; and 'at most one right angle' is the general argument of S12, which no computation on finitely many examples can replace."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "內角量值以 G2 的餘弦式定義；它與綜合幾何中角的量測一致這一點，本稿不證明。",
        "en": "Interior angle measure is defined by the cosine formula of G2; that it agrees with angle measure in synthetic geometry is not proved here."
      },
      {
        "zh-Hant": "cos_facts、line_unique、angle_positive 是前提，未在本稿推導。",
        "en": "cos_facts, line_unique and angle_positive are premises and are not derived in this guide."
      },
      {
        "zh-Hant": "N17（20+5=25）只是與畢氏定理一致的旁證，本稿論證不使用它。",
        "en": "N17 (20+5=25) is only a side check consistent with the Pythagorean theorem; the argument here does not use it."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "59d80db968e3384bccfcb59d56d000c022e245739089f60fee3f366ca873745c",
    "original_source_sha256": "59d80db968e3384bccfcb59d56d000c022e245739089f60fee3f366ca873745c",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}