The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference, of the circle.
任意圆的面积，等于一个直角三角形；这个三角形的一条直角边等于圆半径，另一条直角边等于圆周长。
圆心 O、半径 OA、内接八边形和外切正方形给出阿基米德的基本夹逼图式。内接边界从圆内逼近，外切边界从圆外逼近。
If the circle were greater than the comparison triangle K, keep bisecting arcs and inscribe a polygon until the remaining circular segments are smaller than the excess.
若圆面积大于比较三角形 K，就不断平分圆弧并作内接多边形，使圆与多边形之间剩下的弓形总面积小于这个“多出来”的面积。
The inscribed polygon would then be greater than K, but every triangle from O to one side has height less than the radius and total base less than the circumference.
这样内接多边形会大于 K；但从 O 向每条边分成的小三角形，高小于半径，总底边小于圆周长，所以多边形又小于 K，矛盾。
If the circle were less than K, refine a circumscribed polygon until the outside strips are smaller than the excess of K over the circle.
若圆面积小于 K，就不断作外切多边形，使外切多边形和圆之间的余隙总面积小于 K 超过圆的那一块。
The circumscribed polygon would be less than K, but its apothem is the radius and its perimeter is greater than the circumference, so its area is greater than K.
于是外切多边形会小于 K；但它到每条边的垂距等于半径，周长又大于圆周长，所以面积大于 K，仍然矛盾。
The circle is neither greater nor less than K; therefore it is equal to K.
圆既不能大于 K，也不能小于 K，所以圆面积等于 K。
Modern notation writes this as A = 1/2 rC; when C = 2πr, it becomes A = πr².
现代写法是 A = 1/2 rC；再用 C = 2πr，才得到 A = πr²。这里真正重要的不是公式，而是双重反证的夹逼结构。
不看完整证明，说明“命题 1 · 圆面积与半径-周长直角三角形”这一命题的已知、要证和关键比较对象。
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- 先把几何对象命名，再说它们之间要比较什么量。
- 穷竭法命题要特别留意“若大于”和“若小于”两边。