Pythagorean theorem

    $$A^2+B^2=C^2$$


Proof:

Pythagorean theorem proof
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: