elem.3.35
若圆内两直线相交，则其中一条直线被交点所分两线段所成矩形等于另一条直线被交点所分两线段所成矩形。
本页以“圆内相交弦定理”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 in the circle ABCD let the two straight lines AC, BD cut one another at the point E; I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB. If now AC, BD are through the centre, so that E is the centre of the circle ABCD, it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB. Next let AC, DB not be through the centre; let the centre of ABCD be taken, and let it be F; from F let FG, FH be drawn perpendicular to the straight lines AC, DB, and let FB, FC, FE be joined.
设圆ABCD内两直线AC、BD相交于点E。若AC、BD均过圆心，则E为圆心，AE、EC、DE、EB相等，故所成矩形相等。
Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3] therefore AG is equal to GC. Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E, the rectangle contained by AE, EC together with the square on EG is equal to the square on GC; [II. 5] Let the square on GF be added; therefore the rectangle AE, EC together with the squares on GE, GF is equal to the squares on CG, GF.
若AC、BD不过圆心，取圆心F，作FG⊥AC、FH⊥BD，连接FB、FC、FE。因过圆心的直线GF垂直平分弦AC，故AG=GC。
But the square on FE is equal to the squares on EG, GF, and the square on FC is equal to the squares on CG, GF; [I. 47] therefore the rectangle AE, EC together with the square on FE is equal to the square on FC. And FC is equal to FB; therefore the rectangle AE, EC together with the square on EF is equal to the square on FB. For the same reason, also, the rectangle DE, EB together with the square on FE is equal to the square on FB.
由II.5，AC被G平分、被E分于不等段，则AE·EC+EG²=GC²。两边加GF²，得AE·EC+(EG²+GF²)=CG²+GF²。由I.47，EG²+GF²=EF²，CG²+GF²=FC²，故AE·EC+EF²=FC²。
But the rectangle AE, EC together with the square on FE was also proved equal to the square on FB; therefore the rectangle AE, EC together with the square on FE is equal to the rectangle DE, EB together with the square on FE. Let the square on FE be subtracted from each; therefore the rectangle contained by AE, EC which remains is equal to the rectangle contained by DE, EB.
因FC=FB，故AE·EC+EF²=FB²。同理，DE·EB+EF²=FB²。故AE·EC+EF²=DE·EB+EF²，两边减去EF²得AE·EC=DE·EB。