Goal and scope
In the Euclidean plane represented by ℝ² (with Euclidean length as distance), prove: for every right triangle ABC with the right angle at C, legs of lengths a=d(C,B) and b=d(C,A), and hypotenuse of length c=d(B,A), a²+b²=c². Then check the example C=(1,2), B=(4,6), A=(−7,8) (25+100=125), and use the non-right triangle E=(0,0), F=(4,0), H=(1,2) to show that without a right angle the relation fails in general.
Proves the theorem in the coordinate model of ℝ²: lengths come from the Euclidean length MKO, right angles from the right-triangle MKO, and angle measures from the cosine formula in G2. The proof is the dot-product version, corresponding to the vector-angle version mentioned by the MKO; it does not prove the area-rearrangement argument summarized in the MKO, nor the converse; it does not treat the equivalence of this model with synthetic geometry or with other vertex representations (MKO formal next); and it does not change the canonical MKO or its English edition.
Main derivation
S01 · definition
Before
\text{right triangle }ABC,\ \gamma=\tfrac{\pi}{2}After
a=d(C,B),\ b=d(C,A),\ c=d(B,A);\ \text{goal}\ a^2+b^2=c^2Conditions: none
Rule: def_intro · Uses: G1, G3, mko_length_def, mko_right_triangle_def
Leaf basis: def_intro + mko_right_triangle_def + mko_length_def
Write the data of the theorem as quantities of the model: by the right-triangle MKO, ABC has an interior angle equal to π/2; call its vertex C (vertex names can be permuted freely, so no generality is lost). The legs are the two sides with endpoint C, the hypotenuse is the side opposite C, and their lengths are computed by the Euclidean length MKO, all nonnegative reals (assumption a4). The goal is a²+b²=c².
S02 · definition
Before
B-C,\ A-C
After
u=(u_1,u_2)=B-C,\ v=(v_1,v_2)=A-C,\ u\neq 0,\ v\neq 0
Conditions: none
Rule: def_intro · Uses: G1, G3
Leaf basis: def_intro + coord_ops
Introduce the two side vectors from C: u=B−C and v=A−C. The triangle is nondegenerate, so B≠C and A≠C, hence u and v are nonzero.
S03 · equivalent
Before
A-B
After
v-u
Conditions: none
Rule: coord_ops · Uses: S02
Leaf basis: coord_ops + add_assoc + add_comm + add_inverse
v−u=(A−C)−(B−C)=A−C−B+C=A−B, an identity in each coordinate (N3). So the difference vector along the hypotenuse is expressed through the two leg vectors.
S04 · equivalent
Before
\gamma=\tfrac{\pi}{2}After
u\cdot v=u_1v_1+u_2v_2=0
Conditions: u\neq 0; v\neq 0
Rule: L_right_iff_dot · Uses: S02, G2, L_right_iff_dot
Leaf basis: L_right_iff_dot + angle_cos_def
By G2, the interior angle γ∈[0,π] at C satisfies cos γ=(u·v)/(|u||v|). By Lemma L_right_iff_dot, γ=π/2 ⇔ u·v=0. This turns the geometric 'right angle' into the algebraic 'dot product zero'. The lemma is proved in S04.1 and S04.2.
S05 · equivalent
Before
a=d(C,B),\ b=d(C,A),\ c=d(B,A)
After
a^2=u_1^2+u_2^2,\ b^2=v_1^2+v_2^2,\ c^2=(v_1-u_1)^2+(v_2-u_2)^2
Conditions: a,b,c\ge 0
Rule: sqrt_def · Uses: S02, S03, G1, sqrt_def
Leaf basis: sqrt_def + mko_length_def + eq_replace
By the distance definition, a=√((x_B−x_C)²+(y_B−y_C)²)=√(u₁²+u₂²); the radicand is ≥0 and the principal root satisfies (√s)²=s, so a²=u₁²+u₂². Likewise b²=v₁²+v₂². For c, A−B=v−u (S03), so c²=(v₁−u₁)²+(v₂−u₂)².
S06 · equivalent
Before
(v_1-u_1)^2+(v_2-u_2)^2
After
(v_1^2-2u_1v_1+u_1^2)+(v_2^2-2u_2v_2+u_2^2)
Conditions: none
Rule: L_sq_diff · Uses: S05, L_sq_diff
Leaf basis: L_sq_diff + mul_comm + eq_replace
Apply Lemma L_sq_diff in each coordinate (p=v_i, q=u_i): (v_i−u_i)²=v_i²−2v_iu_i+u_i², and commutativity writes the middle term as −2u_iv_i. The lemma is proved in S06.1.
S07 · equivalent
Before
(v_1^2-2u_1v_1+u_1^2)+(v_2^2-2u_2v_2+u_2^2)
After
c^2=a^2+b^2-2\,u\cdot v
Conditions: none
Rule: add_comm · Uses: S05, S06
Leaf basis: add_comm + add_assoc + distributive + dot_def
Regroup (commutativity and associativity of addition, distributivity): =(u₁²+u₂²)+(v₁²+v₂²)−2(u₁v₁+u₂v₂). By S05 and the definition of the dot product, this is a²+b²−2u·v. Note that this equation holds for every triangle (N1 is a polynomial identity in four variables); the right angle has not been used yet.
S08 · implies
Before
c^2=a^2+b^2-2\,u\cdot v,\ u\cdot v=0
After
a^2+b^2=c^2
Conditions: u\cdot v=0
Rule: eq_replace · Uses: S04, S07
Leaf basis: eq_replace + mul_zero + add_zero + eq_refl_symm_trans
The right angle is used only here: by S04, u·v=0, and substituting into S07 gives c²=a²+b²−2·0=a²+b²; by symmetry of equality, a²+b²=c².
S09 · implies
Before
a^2+b^2=c^2\ \text{for the arbitrary right triangle of G3}After
\text{every right triangle of the model}:\ a^2+b^2=c^2Conditions: none
Rule: forall_intro · Uses: S01, S08
Leaf basis: forall_intro
S01 to S08 use only what G3 says, nondegenerate with a right angle at C, and no particular coordinates, so ABC is arbitrary; by ∀-introduction every right triangle of the model satisfies a²+b²=c². If the right angle is at another vertex, it suffices to rename the vertices.
S10 · implies
Before
C=(1,2),\ B=(4,6),\ A=(-7,8)
After
\triangle ABC\ \text{nondegenerate}Conditions: C\neq B
Rule: line_unique · Uses: G4, line_unique
Leaf basis: line_unique + image_preimage_def + contradiction + numeric_evaluation
The points are pairwise distinct (first coordinates 1, 4, −7 all differ). B−C=(3,4), and A∈L(C,B) ⇔ some t gives (−7,8)=(1+3t, 2+4t) ⇔ t=−8/3 and t=3/2, which is impossible, so A∉L(C,B); by uniqueness of the line the points are noncollinear and the triangle is nondegenerate.
S11 · equivalent
Before
u=B-C=(3,4),\ v=A-C=(-8,6)
After
u\cdot v=-24+24=0,\ \gamma=\tfrac{\pi}{2}Conditions: u\neq 0; v\neq 0
Rule: L_right_iff_dot · Uses: S10, S04, G4
Leaf basis: L_right_iff_dot + dot_def + numeric_evaluation
u=(4−1, 6−2)=(3,4) and v=(−7−1, 8−2)=(−8,6); u·v=3·(−8)+4·6=−24+24=0. By the equivalence S04 the angle at C is right, so the example satisfies G3.
S12 · implies
Before
a^2,\ b^2,\ c^2\ \text{for the example}After
25+100=125
Conditions: none
Rule: numeric_evaluation · Uses: S05, S11, G4
Leaf basis: numeric_evaluation + sqrt_def + power_laws
a²=3²+4²=25 and b²=(−8)²+6²=64+36=100; A−B=(−7−4, 8−6)=(−11,2), so c²=121+4=125; 25+100=125, agreeing with S08. a=5, b=10, c=√125=5√5 (since 5√5≥0 and (5√5)²=25·5=125). This is a check of one example, not a proof of the theorem.
S13 · implies
Before
E=(0,0),\ F=(4,0),\ H=(1,2)
After
d(E,F)^2=16\neq 13+5=d(H,F)^2+d(H,E)^2
Conditions: none
Rule: eq_replace · Uses: S07, G4
Leaf basis: eq_replace + dot_def + numeric_evaluation
A non-right comparison: H is not on the x-axis L(E,F)={(4t,0)} (second coordinate 2≠0), so EFH is nondegenerate. At H: u=F−H=(3,−2), v=E−H=(−1,−2), and u·v=−3+4=1≠0, so the angle at H is not right. d(H,F)²=9+4=13, d(H,E)²=1+4=5, d(E,F)²=16; 13+5=18≠16. The general equation of S07 gives 16=13+5−2·1, showing that what is missing is exactly the term −2u·v. The dot products at E and F are 4 and 12, so those angles are not right either; none of the three pairings 13+5=16, 13+16=5, 5+16=13 holds.
S14 · implies
Before
\text{S08, S09, S12, S13}After
\forall\ \text{right triangles}:\ a^2+b^2=c^2Conditions: none
Rule: and_intro_elim · Uses: S08, S09, S12, S13
Leaf basis: and_intro_elim + forall_intro
The theorem holds in the model: for every right triangle the sum of the squares of the legs equals the square of the hypotenuse. The example 25+100=125 agrees; EFH shows that without a right angle the relation fails in general.