elem.4.8
在给定正方形内作内切圆。
正方形 ABCD；E、F 分别为 AD、AB 的中点；过 E 作 EH 平行于 AB、过 F 作 FK 平行于 AD，二者交于中心 G；以 G 为心、GE 为半径的圆 EFHK 即为内切圆。
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 ABCD be the given square; thus it is required to inscribe a circle in the given square ABCD. Let the straight lines AD, AB be bisected at the points E, F respectively [I. 10], through E let EH be drawn parallel to either AB or CD, and through F let FK be drawn parallel to either AD or BC; [I. 31] therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are evidently equal.
设ABCD为给定正方形，分别取AD和AB的中点E和F。
[I. 34] Now, since AD is equal to AB, and AE is half of AD, and AF half of AB, therefore AE is equal to AF, so that the opposite sides are also equal; therefore FG is equal to GE. Similarly we can prove that each of the straight lines GH, GK is equal to each of the straight lines FG, GE; therefore the four straight lines GE, GF, GH, GK are equal to one another.
过E作AB或CD的平行线EH，过F作AD或BC的平行线FK，得到多个平行四边形，其对边相等。
Therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will pass also through the remaining points. And it will touch the straight lines AB, BC, CD, DA, because the angles at E, F, H, K are right.
由AD等于AB，AE等于AF，可得FG等于GE，同理可证GH、GK均等于FG、GE，故GE、GF、GH、GK四线段相等。
For, if the circle cuts AB, BC, CD, DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle : which was proved absurd; [III. 16] therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will not cut the straight lines AB, BC, CD, DA. Therefore it will touch them, and will have been inscribed in the square ABCD.
以G为圆心、GE为半径作圆，因E、F、H、K处为直角，圆与正方形各边相切，即为所求内切圆。