I have come across many proofs of Pythagoras theorem like these -
But today when i was trying different methods of proving the pythagorean theorem i found a way to prove it. But the construction is so simple and obvious so I wanted to check if this proof is there somewhere -
Consider this right triangle with $\angle B = 90°$
$\Rightarrow △ABC \sim △ADB \Rightarrow \dfrac{AB}{AD}=\dfrac{AC}{AB} \Rightarrow AB^2=AC.AD \cdots (I)$
Similarly ,
$△ABC \sim △BDC \Rightarrow \dfrac{BC}{DC}=\dfrac{AC}{BC} \Rightarrow BC^2=AC.DC \cdots (II)$
Adding $(I) , (II) \Rightarrow AB^2+BC^2=AC(AD+DC)=AC^2$. Hence Proved.
A proof so simple yet I never came across it...