elem.4.2
在一个给定圆内作一个三角形，与给定三角形等角。
圆 ABC 圆心 O，三角形 ABC 内接于圆，A 在顶端；GH 为过 A 的切线；DEF 为给定参照三角形（左下方），所作三角形 ABC 与 DEF 等角。
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 the given circle, and DEF the given triangle; thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF. Let GH be drawn touching the circle ABC at A [III. 16, Por.]; on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23] let BC be joined.
设给定圆为ABC，给定三角形为DEF，需在圆ABC内作三角形与DEF等角。
Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle.
过点A作圆ABC的切线GH，在直线AH上点A处作角HAC等于角DEF，在直线AG上点A处作角GAB等于角DFE。
[III. 32] But the angle HAC is equal to the angle DEF; therefore the angle ABC is also equal to the angle DEF.
连接BC。因切线AH与弦AC所成角HAC等于相对弧上的圆周角ABC，且角HAC等于角DEF，故角ABC等于角DEF。
For the same reason the angle ACB is also equal to the angle DFE; therefore the remaining angle BAC is also equal to the remaining angle EDF.
同理，角ACB等于角DFE，因此剩余角BAC等于剩余角EDF，三角形ABC与DEF等角。