Pythagorean theorem

Proof:

Since $|\hat{a}|=|\hat{d_1}|=90°$ and $|\hat{c}|=|\hat{c}|$ , we can conclude that: $\Delta abc \sim \Delta dac$
This means that:
$\frac{|ab|}{|ad|}=\frac{|bc|}{|ac|}=\frac{|ac|}{|cd|}$
$\frac{|bc|}{|ac|}=\frac{|ac|}{|cd|}$ (Ignore the $\frac{|ab|}{|ad|}$ )
$|ac|^2=|bc||cd|$ (Multiply both sides by $|ac||cd|$ )
$B^2=AB'$

Same exact procedure for: $\Delta abc \sim \Delta dab$ , which means that:
$\frac{|ab|}{|bd|}=\frac{|bc|}{|ab|}$
$|ab|^2=|bc||bd|$ (Multiply both sides by $|ab||bd|$ )
$C^2=AC'$

Now add both $B^2$ and $C^2$ together.
$B^2+C^2=AB'+AC'=A(B'+C')=A^2$ (because $B'+C'=A$ , see picture)

Try it yourself:
x y

Output: