Main derivation
S01 · definition
Before
\triangle ABC\ \text{is a right triangle}After
\triangle ABC\ \text{nondegenerate and some interior angle}=90^\circ=\tfrac{\pi}{2}Conditions: none
Rule: mko_right_triangle_def · Uses: G2, mko_right_triangle_def
Leaf basis: mko_right_triangle_def + def_unfold
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.
S02 · equivalent
Before
90^\circ
After
\tfrac{\pi}{2}\ \text{rad}Conditions: none
Rule: deg_rad · Uses: deg_rad
Leaf basis: deg_rad + frac_rules + numeric_evaluation
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.
S03 · implies
Before
A=(1,1),\ B=(5,3),\ C=(0,3)
After
A\neq B,\ B\neq C,\ C\neq A
Conditions: none
Rule: cartesian_product_def · Uses: G3, cartesian_product_def
Leaf basis: cartesian_product_def + numeric_evaluation
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.
S04 · equivalent
Before
C\in L(A,B)
After
\exists t\in\mathbb{R}:\ t=-\tfrac{1}{4}\ \land\ t=1Conditions: 4\neq 0; 2\neq 0
Rule: image_preimage_def · Uses: G1, G3, S03
Leaf basis: image_preimage_def + cartesian_product_def + eq_congruence + mul_inverse_nonzero
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.
S05 · implies
Before
\exists t:\ t=-\tfrac{1}{4}\ \land\ t=1After
C\notin L(A,B)
Conditions: none
Rule: contradiction · Uses: S04
Leaf basis: contradiction + eq_refl_symm_trans + numeric_evaluation
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).
S06 · implies
Before
A\neq B,\ C\notin L(A,B)
After
\triangle ABC\ \text{nondegenerate}Conditions: A\neq B
Rule: line_unique · Uses: S03, S05, line_unique, mko_triangle_def
Leaf basis: line_unique + collinear_def + contradiction + mko_triangle_def
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.
S07 · equivalent
Before
u=B-A,\ v=C-A
After
u=(4,2),\ v=(-1,2),\ u\cdot v=0
Conditions: none
Rule: dot_def · Uses: G1, G3, S03
Leaf basis: dot_def + coord_ops + numeric_evaluation
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.
S08 · equivalent
Before
u\cdot v=0,\ u\neq 0,\ v\neq 0
After
\theta_A=\tfrac{\pi}{2}Conditions: u\neq 0; v\neq 0
Rule: L_right_iff_dot · Uses: S03, S06, S07, L_right_iff_dot
Leaf basis: L_right_iff_dot + angle_cos_def
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.
S09 · implies
Before
\triangle ABC\ \text{nondegenerate},\ \theta_A=\tfrac{\pi}{2}After
\triangle ABC\ \text{is a right triangle with the right angle at}\ AConditions: none
Rule: mko_right_triangle_def · Uses: S02, S06, S08, mko_right_triangle_def
Leaf basis: mko_right_triangle_def + exists_intro + and_intro_elim
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.
S10 · equivalent
Before
(A-B)\cdot(C-B)=(-4,-2)\cdot(-5,0)
After
20\neq 0,\ \theta_B\neq\tfrac{\pi}{2}Conditions: A-B\neq 0; C-B\neq 0
Rule: L_right_iff_dot · Uses: S03, S06, G3, L_right_iff_dot
Leaf basis: L_right_iff_dot + dot_def + numeric_evaluation
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.
S11 · equivalent
Before
(A-C)\cdot(B-C)=(1,-2)\cdot(5,0)
After
5\neq 0,\ \theta_C\neq\tfrac{\pi}{2}Conditions: A-C\neq 0; B-C\neq 0
Rule: L_right_iff_dot · Uses: S03, S06, G3, L_right_iff_dot
Leaf basis: L_right_iff_dot + dot_def + numeric_evaluation
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.
S12 · implies
Before
\theta_X=\theta_Y=\tfrac{\pi}{2}\ \text{at two vertices}\ X\neq YAfter
\text{contradiction; a nondegenerate triangle has at most one right angle}Conditions: none
Rule: contradiction · Uses: thm_angle_sum, angle_positive
Leaf basis: thm_angle_sum + angle_positive + contradiction + forall_intro
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.
S13 · definition
Before
\theta_A=\tfrac{\pi}{2}After
\text{hypotenuse}\ BC,\ \text{legs}\ AB,\ ACConditions: none
Rule: def_intro · Uses: S09, dot_def
Leaf basis: def_intro + dot_def + numeric_evaluation
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.
S14 · implies
Before
A,\ B,\ D=(1,4)
After
\triangle ABD\ \text{nondegenerate}Conditions: A\neq B
Rule: line_unique · Uses: G1, G3, line_unique
Leaf basis: line_unique + image_preimage_def + contradiction + mko_triangle_def
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.
S15 · implies
Before
(B-A)\cdot(D-A),\ (A-B)\cdot(D-B),\ (A-D)\cdot(B-D)
After
6,\ 14,\ 3\neq 0;\ \triangle ABD\ \text{not right}Conditions: none
Rule: L_right_iff_dot · Uses: S14, L_right_iff_dot, G3
Leaf basis: L_right_iff_dot + dot_def + numeric_evaluation + de_morgan_logic
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.
S16 · implies
Before
\text{S09, S12, S13, S15}After
\triangle ABC\ \text{right at}\ A;\ \triangle ABD\ \text{not right}Conditions: none
Rule: mko_right_triangle_def · Uses: S09, S10, S11, S12, S13, S15
Leaf basis: mko_right_triangle_def + and_intro_elim
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.
Error controls and negative controls
First confirm a nondegenerate triangle
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.
Do not mix degrees and radians
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.
A program's tolerance check is a check of concrete input
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.
- With E=(1,1), (B−A)·(E−A)=0, so ABE is a right triangle with the right angle at A →
reject — 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). - Side AD of triangle ABD is vertical, so ABD has a right angle →
reject — The dot products at the three vertices are 6, 14 and 3, all nonzero (S15), so no interior angle equals π/2. - Some nondegenerate triangles have two right angles →
reject — Then the third angle would be π−π/2−π/2=0, contradicting that every interior angle is positive (S12). - The right angle of ABC is at A, so the hypotenuse is AB →
reject — 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.