elem.3.2
在圆上任取两点，连接这两点的线段必落在圆内。
本页以“圆内弦必在圆内”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。
Let ABC be a circle, and let two points A, B be taken at random on its circumference; I say that the straight line joined from A to B will fall within the circle. For suppose it does not, but, if possible, let it fall outside, as AEB; let the centre of the circle ABC be taken [III. 1], and let it be D; let DA, DB be joined, and let DFE be drawn through. Then, since DA is equal to DB, the angle DAE is also equal to the angle DBE.
设圆ABC，在圆周上任取两点A、B，连接AB。假设AB落在圆外，如AEB。
[I. 5] And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. [I. 16] But the angle DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE.
取圆心D，连接DA、DB，并作DFE穿过。因DA等于DB，故角DAE等于角DBE。
And the greater angle is subtended by the greater side; [I. 19] therefore DB is greater than DE. But DB is equal to DF; therefore DF is greater than DE, the less than the greater : which is impossible.
三角形DAE的边AEB延长，角DEB大于角DAE，从而大于角DBE。大角对大边，故DB大于DE。
Therefore the straight line joined from A to B will not fall outside the circle. Similarly we can prove that neither will it fall on the circumference itself; therefore it will fall within.
但DB等于DF，故DF大于DE，矛盾。因此AB不在圆外，同理也不在圆周上，必在圆内。