三等分器｜陰陽二旋器

很想擁有這個東西，雖然好像也沒甚麼用。

The construction begins with drawing a circle passing through the vertex P of the angle to be trisected, centered at A on an edge of this angle, and having B as its second intersection with the edge. A circle centered at P and of the same radius intersects the line supporting the edge in A and O.

Now the right triangular ruler is placed on the drawing in the following manner: one leg of its right angle passes through O; the vertex of its right angle is placed at a point S on the line PC in such a way that the second leg of the ruler is tangent at E to the circle centered at A. It follows that the original angle is trisected by the line PE, and the line PD perpendicular to SE and passing through P. This line can be drawn either by using again the right triangular ruler, or by using a traditional straightedge and compass construction. With a similar construction, one can improve the location of E, by using that it is the intersection of the line SE and its perpendicular passing through A.

Proof: One has to prove the angle equalities E P D ^ = D P S ^ {\displaystyle {\widehat {EPD}}={\widehat {DPS}}} and B P E ^ = E P D ^ . {\displaystyle {\widehat {BPE}}={\widehat {EPD}}.} The three lines OS, PD, and AE are parallel. As the line segments OP and PA are equal, these three parallel lines delimit two equal segments on every other secant line, and in particular on their common perpendicular SE. Thus SD' = D'E, where D' is the intersection of the lines PD and SE. It follows that the right triangles PD'S and PD'E are congruent, and thus that E P D ^ = D P S ^ , {\displaystyle {\widehat {EPD}}={\widehat {DPS}},} the first desired equality. On the other hand, the triangle PAE is isosceles, since all radiuses of a circle are equal; this implies that A P E ^ = A E P ^ . {\displaystyle {\widehat {APE}}={\widehat {AEP}}.} One has also A E P ^ = E P D ^ , {\displaystyle {\widehat {AEP}}={\widehat {EPD}},} since these two angles are alternate angles of a transversal to two parallel lines. This proves the second desired equality, and thus the correctness of the construction.

此構造始於繪製一個通過待三等分角之頂點 P 的圓，該圓以該角一邊上的 A 點為圓心，且與該邊的第二個交點為 B。一個以 P 為圓心，且半徑相同的圓與該邊所在的直線相交於 A 和 O 點。

現在，將直角三角板以下列方式放置在圖上：其直角的一條邊通過 O 點；其直角的頂點放置在直線 PC 上的一點 S，使得三角板的第二條邊與以 A 為圓心的圓相切於 E 點。由此可知，原始角被直線 PE 和垂直於 SE 且通過 P 點的直線 PD 三等分。這條直線可以使用直角三角板再次繪製，或者使用傳統的直尺和圓規構造。透過類似的構造，可以透過使用 SE 直線及其通過 A 點的垂線的交點來改善 E 點的位置。

證明：必須證明角度相等 E P D ^ = D P S ^ {\displaystyle {\widehat {EPD}}={\widehat {DPS}}} 和 B P E ^ = E P D ^ 。 {\displaystyle {\widehat {BPE}}={\widehat {EPD}}.} 三條直線 OS、PD 和 AE 平行。由於線段 OP 和 PA 相等，因此這三條平行線在每條割線上劃分出兩個相等的線段，特別是在它們的共同垂線 SE 上。因此 SD' = D'E，其中 D' 是直線 PD 和 SE 的交點。由此可知，直角三角形 PD'S 和 PD'E 全等，因此 E P D ^ = D P S ^ , {\displaystyle {\widehat {EPD}}={\widehat {DPS}},} 即第一個期望的等式。另一方面，三角形 PAE 是等腰三角形，因為圓的所有半徑都相等；這意味著 A P E ^ = A E P ^ 。 {\displaystyle {\widehat {APE}}={\widehat {AEP}}.} 還有 A E P ^ = E P D ^ , {\displaystyle {\widehat {AEP}}={\widehat {EPD}},} 因為這兩個角是橫截線與兩條平行線的內錯角。這證明了第二個期望的等式，從而證明了構造的正確性。