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|CB| = |CA| + |AB|
2 = |CA| + r
|CA| = 2 - r
z tw. pitagorasa:
r² = 1² + |CA|²
r² = 1² + (2 - r)²
r² - (2 - r)² = 1²
(r - 2 + r)(r + 2 - r) = 1
(2r - 2)2 = 1
4r - 4 = 1
r = 5/4
trójkąt na rysunku