English: Simplified version of similar triangles proof for Pythagoras' theorem.

In triangle ACB, angle ACB is the right angle. CH is a perpendicular on hypotenuse AB of triangle ACB.
In triangle AHC and triangle ACB, ∠AHC=∠ACB as each is a right angle. ∠HAC=∠CAB as they are common angles at vertex A. Thus triangle AHC is similar to triangle ACB by AA test.
Thus, ${\displaystyle {\frac {AC}{AB}}={\frac {AH}{AC}}}$   ${\displaystyle \therefore {AC}^{2}={AB}\times {AH}}$

In triangle BHC and triangle ACB, ∠BHC=∠ACB as each is a right angle. ∠HBC=∠CBA as they are common angles at vertex B. Thus triangle BHC is similar to triangle BCA by AA test.
Thus, ${\displaystyle {\frac {BC}{AB}}={\frac {BH}{BC}}}$   ${\displaystyle \therefore {BC}^{2}={AB}\times {BH}}$

${\displaystyle \therefore {AC}^{2}+{BC}^{2}=({AB}\times {AH})+({AB}\times {BH})}$

${\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}\times ({AH}+{BH})}$

${\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}\times {AB}}$

${\displaystyle \therefore {AC}^{2}+{BC}^{2}={AB}^{2}}$ which is the Pythagoras theorem.