In impedance triangle ABC, in which quadrant is angle g, is: | |
            
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Mathematical evidence by Law of cosine:        As evidence of Pythagoras' theorem is enough to use equation: c2 = a2 + b2 - 2 a b cos g    from Law of cosine and substitute g = p / 2 . Evidence of inverted Pythagoras' theorem:        We'll make evidence by dissension: Let's predict that c2 = a2 + b2, while angle opossite to side c isn't quadrant. We can construct triangle ABC1 so that:        |AC1| = |AC|,        |angle AC1B| = 90o ![]()        a12 + b2 = a2 + b2 We'll substract b2 : a12 = a2        a1 and a are positive then we have: a1 = a - triangles ABC and ABC1 are identical by sss (side, side, side) but aren't identical in matching included angles ACB and AC1 - dissension. Assumption that angle ACB is quadrant is wrong. That means that if c2 = a2 + b2 is true then this triangle is impedance with quadrant opposite to side c - QED. Evidence by similarity : We'll use Euklides' theorem:        ![]() a2 + b2 = c.ca + c.cb = c ( ca + cb ) = c2 - > c2 = a2 + b2. |