In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.
在钝角三角形中，钝角所对边上的正方形，大于夹钝角两边上正方形之和，超出量等于两倍某边与投影段所成矩形。
钝角三角形 ABC，顶点 B 为钝角；过 A 作 AD⊥BC，垂足 D 落在 CB 沿 B 反向的延长线上。CD² = CB² + BA² + 2·CB·BD。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and let BD be drawn from the point B perpendicular to CA produced; I say that the square on BC is greater than the squares on BA, AC by twice the rectangle contained by CA, AD.
在钝角三角形中，从一锐角顶点向钝角邻边延长线作垂线。
For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD.
斜边被表示为邻边加投影段。
[II. 4] Let the square on DB be added to each; therefore the squares on CD, DB are equal to the squares on CA, AD, DB and twice the rectangle CA, AD.
用 euclid-elements/book2-prop-004 展开这条和线上的正方形。
But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47] and the square on AB is equal to the squares on AD, DB; [I. 47] therefore the square on CB is equal to the squares on CA, AB and twice the rectangle contained by CA, AD; so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD.
再用直角三角形中的 euclid-elements/book1-prop-047 替换垂线相关平方，得到超出量为两倍矩形。