Unit V: Synthetic Appendix (Anhang) — Pure Euclidean Proofs of Nine Theorems (pp. 56–62)¶
Karl Wilhelm Feuerbach — Properties of Several Remarkable Points of the Rectilinear Triangle (1822, pp. 56–62, Fig. 13–17)
Synopsis of Propositions 1–4 (pp. 56–59)¶
In the closing Anhang, Feuerbach demonstrates the foundational metric lemmas of Sections II and III purely synthetically from Books I–IV and VI of Euclid’s Elements, without trigonometric functions or coordinate algebra:
1st Proposition (*1. Satz*): In any triangle \(ABC\) with circumcenter \(K\), orthocenter \(O\), and midpoint \(D\) of side \(AB\), the altitude segment from the opposite vertex equals twice the distance of the circumcenter from that side: \(CO = 2\,KD\).
2nd Proposition (*2. Satz*): The square on any side \(AB\) added to the square on the opposite vertex-to-orthocenter segment \(CO\) equals the square on the diameter of the circumscribed circle: \(\overline{AB}^2 + \overline{CO}^2 = 4R^2\).
3rd & 4th Propositions (*3.–4. Satz*): From cyclic orthic quadrilaterals and [VI. 16], the rectangle contained by the inradius \(r\) and the perimeter of \(\triangle ABC\) equals the rectangle contained by the circumradius \(R\) and the perimeter of the orthic triangle \(\triangle MNP\):
\[r(AB + AC + BC) = R(MN + MP + NP) \qquad \Longleftrightarrow \qquad MN + MP + NP = \frac{2\Delta}{R}.\]
5th Proposition (5. Satz) — Ratio of the Orthic Area to ΔABC (Complete Translation)¶
Enunciation. The area of triangle \(MNP\), which the feet of the altitudes form in an acute-angled triangle \(ABC\), bears to the area of triangle \(ABC\) the same ratio as the radius of the circle inscribed in the former bears to the radius of the circle circumscribed about the latter.
Proof. For if the radius of the circle inscribed in triangle \(MNP\) is \(\rho\), and \(r, R\) are as before the radii of the circles inscribed in and circumscribed about triangle \(ABC\), then as is well known \(2\,\triangle MNP = \rho(MN + MP + NP)\) and \(2\,\triangle ABC = r(AB + AC + BC)\), whence
Now however there follows from the preceding [4th] Proposition [VI. 16] \(r(AB + AC + BC) = R(MN + MP + NP)\), and therefore:
6th Proposition (6. Satz) — Altitude-Segment Rectangle CO · OP = 2Rρ¶
Enunciation. The intersection point \(O\) of the altitudes in an acute-angled triangle \(ABC\) divides each of them \(CP\) into two such segments \(CO, OP\) that the rectangle contained by them equals the rectangle contained by the diameter of the circle circumscribed about triangle \(ABC\) and the radius of the circle inscribed in the triangle \(MNP\) formed by the feet of those altitudes.
Proof. Erect from \(O\) upon side \(MP\) the perpendicular \(OJ\). Because [III. 22] a circle can be circumscribed about the quadrilateral \(ANOP\), we have [III. 21] \(\angle NPO = \angle NAO\), and for the very same reason \(\angle MPO = \angle MBO\). Because however [I. 15] \(\angle AON = \angle BOM\) and \(\angle ANB = \text{right} = \angle AMB\), we have \(\angle NAO = \angle MBO\), consequently also \(\angle NPO = \angle MPO\). In just the same way it is proved that \(\angle NMO = \angle PMO\); therefore [IV. 4] \(O\) is the center of the circle inscribed in triangle \(MNP\) and the perpendicular \(OJ\) is a radius [\(\rho\)] of this circle.
Now pass through the vertex \(A\) and the center \(K\) of the circle circumscribed about triangle \(ABC\) a diameter \(AG = 2R\) of the same, and draw \(BG\); then [III. 31] \(\angle ABG = \text{right}\) and [III. 21] \(\angle AGB = \angle ACB\), consequently \(\angle GAB = \angle CBN\). Since now \(\angle MBO = \angle MPO\), we also have \(\angle GAB = \angle MPO\), and hence the right triangles \(ABG, OPJ\) are equiangular, so that [VI. 4] \(AG : BG = OP : OJ\).
Erect furthermore from \(K\) upon \(AB\) the perpendicular \(KD\); then triangles \(ADK, ABG\) are equiangular, whence [VI. 4] \(AD : DK = AB : BG\), and because [III. 3] \(AB = 2\,AD\), we also have \(BG = 2\,KD\). By the 1st Proposition, however, \(CO = 2\,KD\), consequently \(BG = CO\); and therefore \(AG : CO = OP : OJ\) and [VI. 16]:
7th Proposition (7. Satz) — Sum of Squared Sides AB² + AC² + BC² = 8R² + 4ρR¶
Enunciation. The sum of the squares on the sides of an acute-angled triangle \(ABC\) equals twice the square on the diameter of the circumscribed circle, increased by the rectangle contained by this diameter and the diameter of the circle inscribed in triangle \(MNP\) formed by the feet of the altitudes in triangle \(ABC\).
Proof. Let \(O\) be the intersection point of the altitudes. Draw from any vertex \(C\) of triangle \(ABC\) to the midpoint \(D\) of the opposite side \(AB\) the median \(CD\). Since \(AB\) is bisected at \(D\), we have [II. 5] \(AP \cdot BP + \overline{PD}^2 = \frac{1}{4}\overline{AB}^2\). Because however \(\angle ABN = \angle ACP\), the right triangles \(APC, BPO\) are equiangular, whence [VI. 4] \(AP : CP = OP : BP\) and [VI. 16] \(AP \cdot BP = CP \cdot OP\). Since now \(CP = CO + OP\), we have also [II. 3] \(AP \cdot BP = CO \cdot OP + \overline{OP}^2\), and therefore \(CO \cdot OP + \overline{OP}^2 + \overline{PD}^2 = \frac{1}{4}\overline{AB}^2\).
Now by [I. 47] and [II. 4], \(\overline{CD}^2 = \overline{CO}^2 + 2\,CO \cdot OP + \overline{OP}^2 + \overline{PD}^2 = \frac{1}{4}\overline{AB}^2 + \overline{CO}^2 + CO \cdot OP\). Furthermore, by a well-known corollary to [II. 12] [Apollonius’s median theorem]:
and adding \(\overline{AB}^2\) to both sides gives \(\overline{AB}^2 + \overline{AC}^2 + \overline{BC}^2 = 2\overline{AB}^2 + 2\overline{CO}^2 + 2\,CO \cdot OP\). Denoting the radii of the circles circumscribed about \(\triangle ABC\) and inscribed in \(\triangle MNP\) by \(R, \rho\), we have by the 2nd Proposition \(\overline{AB}^2 + \overline{CO}^2 = 4R^2\) and by the 6th Proposition \(CO \cdot OP = 2\rho R\). Therefore:
8th Proposition (8. Satz) — Pedal Segment GH Equals Semiperimeter of ΔMNP¶
Enunciation. When from any altitude foot \(P\) in side \(AB\) of an acute-angled triangle \(ABC\) one drops upon the other two sides \(AC, BC\) the perpendiculars \(PG, PH\) (Fig. 16), the straight line \(GH\) joining the feet of these perpendiculars equals the semiperimeter of the orthic triangle \(MNP\).
Proof. Produce \(PG, PH\) until they meet the prolongation of \(MN\) at \(J\) and \(K\). Because [III. 22] a circle can be circumscribed about quadrilateral \(CNOM\), we have [III. 21] \(\angle CMN = \angle CON\); likewise \(\angle BMP = \angle BOP\); consequently [I. 15] \(\angle CMN = \angle BMP\) and hence also \(\angle BMK = \angle BMP\). Now \(PH \perp BC\), whence [I. 26] \(MP = MK\) and \(PH = HK\). In just the same way it is proved that \(NP = NJ\) and \(PG = GJ\).
Since therefore \(G, H\) are the midpoints of \(PJ, PK\), [VI. 2] \(GH \parallel JK\), whence [VI. 4] \(PG : GH = PJ : JK\). Now \(PJ = 2\,PG\), so also \(JK = 2\,GH\); and because \(JK = MN + MK + NJ = MN + MP + NP\), we have:
9th Proposition (9. Satz) — Pure Euclidean Proof of the Nine-Point Circle¶
Enunciation. The radius of the circle circumscribed about triangle \(MNP\), which the feet of the altitudes form in triangle \(ABC\), is half as great as the radius of the circle circumscribed about triangle \(ABC\); and the locus \(L\) of the center of the former circle bisects the distance of the center \(K\) of the latter circle from the intersection point \(O\) of those altitudes (Fig. 17).
Proof. Draw from \(K\) [X3] to \(O\) [X4] and \(C\) the straight lines \(OK, CK\), and upon \(AB\) drop the perpendicular \(KD\) [to the side midpoint D]. Furthermore pass through \(D\) and the midpoint \(L\) [X5] of the straight line \(OK\) the line \(DL\), which prolonged meets the altitude \(CP\) at \(J\) [the Euler midpoint J].
Because \(LO = LK\), and furthermore [I. 15] \(\angle OLJ = \angle KLD\) and—since \(OJ \parallel KD\)—[I. 29] \(\angle LOJ = \angle LKD\), it follows [I. 26] that \(OJ = KD\) and \(JL = DL\). Now since by the 1st Proposition \(KD = \frac{1}{2}CO\), we also have \(OJ = \frac{1}{2}CO\) [so \(J\) is the Euler midpoint of \(CO\)!] and therefore \(CJ = KD\); and because \(CJ \parallel KD\), we have [I. 33] \(JD = CK\), consequently:
Drop furthermore from \(L\) upon \(AB\) the perpendicular \(LE\) and draw \(LP\) [to the altitude foot P]; because the straight lines \(OP, LE, KD\) all run parallel to one another, [VI. 2] \(OL : KL = PE : DE\), and since \(OL = KL\), we also have \(PE = DE\); hence [I. 4] \(LD = LP\). Since now \(LD = \frac{1}{2}CK\), we also have \(LP = \frac{1}{2}CK\).
In the very same manner it is proved that \(LN = \frac{1}{2}BK\) and \(LM = \frac{1}{2}AK\); consequently, because \(AK = BK = CK\) [\(= R\)], we have also:
and \(L\), that is, the midpoint of \(OK\), is the center of the circle circumscribed about triangle \(MNP\). Q.E.D.