elem.3.24
在相等线段上的相似圆弓形彼此相等。
本页以“等圆上相似弓形相等”整体图解辅助阅读；点、线、角、圆索引已按命题文字和证明步骤校订，可与证明和问答联动。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
For let AEB, CFD be similar segments of circles on equal straight lines AB, CD; I say that the segment AEB is equal to the segment CFD.
设AEB和CFD是在相等线段AB、CD上的相似圆弓形。
For, if the segment AEB be applied to CFD, and if the point A be placed on C and the straight line AB on CD, the point B will also coincide with the point D, because AB is equal to CD; and, AB coinciding with CD, the segment AEB will also coincide with CFD.
将弓形AEB叠合到CFD上，使点A落在C上，线段AB落在CD上，则点B与D重合，因为AB等于CD。
For, if the straight line AB coincide with CD but the segment AEB do not coincide with CFD, it will either fall with it, or outside it; or it will fall awry, as CGD, and a circle cuts a circle at more points than two : which is impossible.
若AB与CD重合而弓形AEB不与CFD重合，则它要么落在其内或外，要么歪斜如CGD，这样圆与圆相交多于两点，这是不可能的（III.10）。
[III. 10] Therefore, if the straight line AB be applied to CD, the segment AEB will not fail to coincide with CFD also; therefore it will coincide with it and will be equal to it.
因此，当AB与CD重合时，弓形AEB必与CFD重合，从而相等。