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 = ABBC/AC we derive (AB^2/AC)AB + ABAC = (ABBC/AC)BC. Dividing by AB/AC leads to AB^2 + AC^2 = BC^2.