Rachel I. Mason's 1996 Proof

Given a right triangle \( \bigtriangleup ABC \) with \( \angkle = \theta \) and \( \angle C = 90^{\circ} \). Let \( D \) be the perpendicular foot from \( C \) to \( \overleftrightarrow{AB} \). Please see the diagram below.

It is easy to see the following: \[ \overline{AD}=b\cdot\cos(\theta),\mbox{\ \ \ \ \ \ \ \ \ \ } \overline{BD}=a\cdot\sin(\theta)\mbox{\ \ \ \ \ a\ \ \ \ \ nd\ \ \ \ \ \ \ \ \ \ } \overline{CD}=d=a\cdot\cos(\theta)=b\cdot\sin(\theta) \] Because \( c=\frac{b}{\cos(\theta)}\) and \(c=\frac{a}{\sin(\theta)}\), adding both together gives \[ 2c = \frac{b\cdot\sin(\theta)+a\cdot\cos(\theta)}{\sin(\theta)\cos(\theta)} \] Substituting \( c = b\cdot\cos(\theta) + a\sin(\theta)\) into the above and rearranging gives the following: \[ b\cdot\sin(\theta)\left( 2\cos^2(\theta)-1\right)+a\cdot\cos(\theta)\left(2\sin^2(\theta)-1\right)=0 \] Because \( b\cdot\sin(\theta)=a\cdot\cos(\theta)\neq 0 \) if \(\theta \neq 0 \), after simplifying yields the Pythagorean Identity.

It is interesting to note that when Mason preparing her paper, she was a high school girl at Framingham High School (please see the image below). Mason's proof of the Pythagorean Identity was 13 years earlier than Zimba's 2009 Proof and 27 years earlier than Jackson-Johnson's 2023 Impossible proof. So, the title of offering the first impossible proof should go to Rachel I. Mason, although I am sure there could be other even earlier proofs. However, please do keep in mind that Rachel I. Mason was a high school girl when she did the proof.

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