Proof By Area method

Draw three squares using the sides of the right triangle ABC as one of their sides, respectively.

The altitude from the right angle ∠ACB intersects the hypotenuse AB at point J, and the side HI of square ABHI at point K.

Connect point A to G and point C to H

Look at △ABG and △CBH ,

∠ABG =∠CBH = 90° + ∠ABC\kappa

According to the condition of triangles congruent SAS,

△ABG \cong △CBH

In the right trapezoid ABCF,

S_{CBGF} = 2S_{△ABG}

In the right trapezoid CBHK,

S_{BHKJ} = 2S_{△CBH}

\therefore S_{CBGF} =S_{BHKJ}

Similarly,

S_{ACED} = S_{AJKI}

\therefore S_{CBGF} + S_{ACED} = S_{BHKJ} + S_{AJKI} = S_{ABHI}

\therefore a^2+b^2 = c^2