Commentary on Proposition IV.4¶
Back to Proposition IV.4.
Euclid does not think it necessary to prove that BD, CD will meet; this is indeed obvious, for the angles DBC, DCB are together half of the angles ABC, ACB, which themselves are together less than two right angles, and therefore the two bisectors of the angles B, C must meet, by Post. 5.
It follows from the proof of this proposition that, if the bisectors of two angles B, C of a triangle meet in D, the line joining D to A also bisects the third angle A, or the bisectors of the three angles of a triangle meet in a point.
It will be observed that Euclid uses the indirect form of proof when showing that the circle touches the three sides of the triangle. Simson proves it directly, and points out that Euclid does the same in III. 17, 33 and 37, whereas in IV. 8 and 13 as well as here he uses the indirect form. The difference is unimportant, being one of form and not of substance; the indirect proof refers back to III. 16, whereas the direct refers back to the Porism to that proposition.
We may state this problem in the more general form: To describe a circle touching three given straight lines which do not all meet in one point, and of which not more than two are parallel.
In the case (1) where two of the straight lines are parallel and the third cuts them, two pairs of interior angles are formed, one on each side of the third straight line. If we bisect each of the interior angles on one side, the bisectors will meet in a point, and this point will be the centre of a circle which can be drawn touching each of the three straight lines, its radius being the perpendicular from the point on any one of the three. Since the alternate angles are equal, two equal circles can be drawn in this manner satisfying the given condition.
In the case (2) where the three straight lines form a triangle, suppose each straight line produced indefinitely. Then each straight line will make two pairs of interior angles with the other two, one pair forming two angles of the triangle, and the other pair being their supplements. By bisecting each angle of either pair we obtain, in the manner of the proposition, two circles satisfying the conditions, one of them being the inscribed circle of the triangle and the other being a circle escribed to it, i.e. touching one side and the other two sides produced. Next, taking the pairs of interior angles formed by a second side with the other two produced indefinitely, we get two circles satisfying the conditions, one of which is the same inscribed circle that we had before, while the other is a second escribed circle. Similarly with the third side. Hence we have the inscribed circle, and three escribed circles (one opposite each angle of the triangle), i.e. four circles in all, satisfying the conditions of the problem.
It may perhaps not be inappropriate to give at this point Heron’s elegant proof of the formula for the area of a triangle in terms of the sides, which we usually write thus:
$$\Delta = \sqrt{s(s - a)(s - b)(s - c)},$$
although it requires the theory of proportions and uses some ungeometrical expressions, e.g. the product of two areas and the “side” of such a product, where of course the areas are so many square units of length. The proof is given in the Metrica, I. 8, and in the Dioptra, 30 (Heron, Vol. III., Teubner, 1903, pp. 20—24 and pp. 280—4, or Heron, ed. Hultsch, pp. 235—7).
Suppose the sides of the triangle ABC to be given in length.
Inscribe the circle DEF, and let G be its centre.
Join AG, BG, CG, DG, EG, FG.
Then
$$\begin{aligned} BC \cdot EG &= 2 \cdot \Delta BGC, \\ CA \cdot FG &= 2 \cdot \Delta ACG, \\ AB \cdot DG &= 2 \cdot \Delta ABG. \end{aligned}$$
Therefore, by addition,
$$p \cdot EG = 2 \cdot \Delta ABC,$$
where $p$ is the perimeter.
Produce CB to H, so that BH = AD.
Then, since AD = AF, DB = BE, FC = CE,
$$CH = \tfrac{1}{2}p.$$
Hence
$$CH \cdot EG = \Delta ABC.$$
But CH · EG is the “side” of the product $CH^2 \cdot EG^2$, that is $\sqrt{CH^2 \cdot EG^2}$; therefore
$$(\Delta ABC)^2 = CH^2 \cdot EG^2.$$
Draw GL at right angles to CG, and BL at right angles to CB, meeting at L. Join CL.
Then, since each of the angles CGL, CBL is right, CGBL is a quadrilateral in a circle.
Therefore the angles CGB, CLB are equal to two right angles.
Now the angles CGB, AGD are equal to two right angles, since AG, BG, CG bisect the angles at G, and the angles CGB, AGD are equal to the angles AGC, DGB, while the sum of all four is equal to four right angles.
Therefore the angles AGD, CLB are equal.
So are the right angles ADG, CBL.
Therefore the triangles AGD, CLB are similar.
Hence
$$\begin{aligned} BC : BL &= AD : DG \\ &= BH : EG, \end{aligned}$$
and, alternately,
$$\begin{aligned} CB : BH &= BL : EG \\ &= BK : KE, \end{aligned}$$
whence, componendo,
$$CH : HB = BE : EK.$$
It follows that
$$\begin{aligned} CH^2 : CH \cdot HB &= BE \cdot EC : CE \cdot EK \\ &= BE \cdot EC : EG^2 \end{aligned}$$
Therefore
(Delta ABC)^2 = CH^2 cdot EG^2 = CH cdot HB cdot CE cdot EB = tfrac{1}{2}p,(tfrac{1}{2}p - BC)(tfrac{1}{2}p - AB)(tfrac{1}{2}p - AC).