How to prove: On the small side AB add a right-angled triangle ABD similar to ABC. Then, naturally, DBC is similar to the other two. From area(ABD) + area(ABC) = area(DBC), AD = AB^2/AC and BD = AB·BC/AC we derive (AB^2/AC)·AB + AB·AC = (AB·BC/AC)·BC. Dividing by AB/AC leads to AB^2 + AC^2 = BC^2.
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