$a^2+b^2=c^2$ ### Rearrangement Proof ![[pythagorean-theorem-rearrangement-proof.svg]] ### Similar triangles proof $ \frac{B C}{A B}=\frac{B H}{B C} \text { and } \frac{A C}{A B}=\frac{A H}{A C} \text {. } $ The first result equates the [[Cosine Function|cosines]] of the angles $\theta$, whereas the second result equates their sines. These ratios can be written as $ B C^2=A B \times B H \text { and } A C^2=A B \times A H \text {. } $ Summing these two equalities results in $ B C^2+A C^2=A B \times B H+A B \times A H=A B(A H+B H)=A B^2, $ which, after simplification, demonstrates the Pythagorean theorem: $ B C^2+A C^2=A B^2 . $