This proof is a variation on #6. 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 = AB²/AC and BD = AB•BC/AC we derive (AB²/AC)•AB + AB•AC = (AB•BC/AC)•BC. Dividing by AB/AC leads toAB² + AC² = BC².