Commentary on Proposition IV.3¶
Back to Proposition IV.3.
The remarks as to the number of ways in which Prop. 2 can be solved apply here also.
Euclid leaves us to satisfy ourselves that the three tangents will meet and form a triangle. This follows easily from the fact that each of the angles AKB, BKC, CKA is less than two right angles. The first two are so by construction, being the supplements of two angles of the given triangle respectively, and, since all three angles round K are together equal to four right angles, it follows that the third, the angle AKC, is equal to the sum of the two angles E, F of the triangle, i.e. to the supplement of the angle D, and is therefore less than two right angles.
Peletarius and Borelli gave an alternative solution, first inscribing a triangle equiangular to the given triangle, by IV. 2, and then drawing tangents to the circle parallel to the sides of the inscribed triangle respectively. This method will of course give two solutions, since two tangents can be drawn parallel to each of the sides of the inscribed triangle.
If the three pairs of parallel tangents be drawn and produced far enough, they will form eight triangles, two of which are the triangles circumscribed to the circle in the manner required in the proposition. The other six triangles are so related to the circle that the circle touches two of the sides in each produced, i.e. the circle is an escribed circle to each of the six triangles.