Chapter 4. Feuerbach's Theorem Let A be a point in the plane and k a positive number. Then in the previous chapter we proved that the inversion mapping with centre A and radius k is the mapping Inv : PnfAg ! PnfAg which is de¯ned as follows. If B1 is a point, then Inv(B1) = B2 if B2 lies on the line joining A and B1 and 2 jAB1jjAB2j = k : We denote this mapping by Inv(A; k2). We proved the following four prop- erties of the mapping Inv(A; k2). (a) If A belongs to a circle C(O; r) with centre O and radius r, then Inv(C(O; r)) is a line l which is perpendicular to OA. (b) If l is a line which does not pass through A, then Inv(l) is a circle such that l is perpendicular to the line joining A to the centre of the circle. (c) If A does not belong to a circle C(O; r), then Inv(C(O; r)) = C(O; r) k2 with r0 = r: ½(A; C(O; r)) (d) If Inv(B1) = B2 and Inv(C1) = C2; where B1 and C1 are two points in the plane, then 1 k2 jB2C2j = jB1C1j: : jAB1jjAC1j Remark Let A be an arbitrary point which does not belong to the circle C(O; r) with centre O and radius r and let ½C(A) be the power of A with respect to the circle C = C(O; r). Then if Inv is the mapping with pole 2 (centre) A and k = ½C(A); i.e. Inv := Inv(A; ½C(A)) then Inv(C(O; r)) = C(O; r); i.e. C(O; r) is invariant under the mapping Inv. This follows from the following observations. Since A does not belong to C(O; r); then Inv(C(O; r)) is a circle with radius r0 where k2 r0 = r ; by (c) above ½C(A) 2 = r; since k = ½C(A) Furthermore, if P is any point of C(O; r) and P 0 = Inv(P ); then 0 jAP jjAP j = ½C(A): Thus P 0 is also on the circle C(O; r); so the result follows. Feuerbach's Theorem The nine point circle of a triangle is tan- gent to the incircle and the three excircles of the triangle. We prove this using inversion. The proof is developed through a sequence of steps. Step 1 Let ABC be a triangle and let Inv be the mapping Inv(A; k2) for some k > 0. If C(O; R) denotes the circumcircle of ABC, then Inv(C(ABC)) is a line L which is antiparallel to the line BC. 2 Proof Let O be the circumcentre of the tri- angle ABC and let Inv denote the mapping Inv(A; k2). From part (a) of the proposition listing the proper- ties of inversion maps, Inv(C(ABC)) = B1C1 where B1;C1 are images of B and C under Inv: Then the line through B1 and C1 is perpendicular to line AO (Figure 1). Now let TA be tangent to C(ABC) at A. Then if X is the point of intersection of AO with C(ABC); we have Figure 1: T ABb = 90± ¡ BAXb = BXAb = BCAb . Since TA?AO and AO?B0C0, then TAkB0C0 so T ABb = ABc0C0: Thus ABc0C0 = BCAb and so B0C0 is antiparallel to BC, as desired. Step 2 Let ABC be a triangle. The incircle C(I; r); with centre I and radius r, touches the sides BC; CA and AB at the points P; Q and R respectively (Figure 2). If s = 1 (a + b + c) denotes the semiperimeter, we have 2 jCP j = jCQj = s ¡ c, jBP j = jBRj = s ¡ b, jARj = jAQj = s ¡ a. Proof Let Figure 2: x = jCP j = jCQj, y = jARj = jAQj, x = jBRj = jBP j. Then s = x + y + z and a = x + z; b = x + y and c = y + z: 3 So jCP j = jCQj = x = (x + y + z) ¡ (y + z) = s ¡ c. Similarly for the other lengths. Step 3 Let ABC be a triangle and let C(Ia; ra) be the excircle touching the side BC and the sides AB and AC externally at the points Pa;Ra and Qa respectively (Figure 3). Then jBPaj = s ¡ c and jCPaj = s ¡ b. Proof Let jARaj = jAQaj = x, jCQaj = jCPaj = y, jBPaj = jBRaj = z. Then x ¡ y = b, x ¡ z = c, y + z = a. Figure 3: Adding, we get 2x = a + b + c so x = s: From this y = x ¡ b = s ¡ b, so jCPaj = jCQaj = s ¡ b, and z = x ¡ c = s ¡ c, so jBPaj = jBRaj = s ¡ c; as required. Figure 4: Remark If A1 is the midpoint of the side BC; then 4 jA1P j = jA1Paj (Figure 4). This follows from the observation that jBP j = s ¡ b (Step 2) jCPaj = s ¡ b (Step 3) a b ¡ c Then jA P j = ¡ (s ¡ b) = ; 1 2 2 a b ¡ c jA P j = ¡ (s ¡ b) = : 1 a 2 2 Step 4 If ABC is a triangle and A3 is a point on the side BC where the bisector of the angle at A meets BC (Figure 5), then ac ab jBA j = and jCA j = : 3 b + c 3 b + c Ab jABjjAA3j sin( ) jBA3j area(ABA3) 2 Proof = = = Figure 5: jCA3j area(ACA3) Ab jACjjAA j sin( ) 3 2 jABj a = : jACj b jBA j c Since jBA j + jCA j = a; then 3 = : 3 3 a ¡ jBA j b ac 3 Solve for jBA j to get jBA j = : 3 3 b + c ac ab Finally, jCA j = a ¡ = : 3 b + c b + c Step 5 In a triangle ABC let A1 be the midpoint of the side BC and let A2 be the foot of the perpendicular from A to BC (Figure 6). Then b2 ¡ c2 jA A j = : 1 2 2a 5 Figure 6: Proof We have 2 2 2 2 2 2 b ¡ c = jAA2j + jA2Cj ¡ jAA2j ¡ jA2Bj = (jA2Cj + jA2Bj)(jA2Cj ¡ jA2Bj) = afjA1A2j + jA1Cj ¡ jA1Bj + jA1A2jg = 2ajA1A2j: b2 ¡ c2 Thus jA A j = ; as required. 1 2 2a Figure 7: Step 6 Let C(O; r) be a circle with centre O and radius r and let A be an arbitrary point not belonging to C(O; r): Con- 2 sider the inversion with pole A and k = ½C(A); the power of A with respect to the circle C(O; r): Then the circle C(O; r) remains invariant under the inversion Inv(A; ½C(A)): Proof Denote by Inv the inversion Inv(A; ½C(A)): Since A does not belong to C(O; r); then Inv(C(O; r)) is a circle with radius r0 where k2 r0 = r: = r; ½C 2 since we have k = ½C(A): Now choose a point B on C(O; r) and let B0 = Inv(B): Then jABjjAB0j = 2 0 k = ½C(A): But this implies that B is a point of C(O; r): Thus Inv(C(O; r)) = C(O; r); as required. 6 Step 7 In a triangle ABC let A1;A2;A3 and P be the following points on the side BC; A1 is the midpoint, A2 is the foot of the altitude from A, b A3 is the point where the bisector of A meets BC, P is the foot of the perpendicular from the incentre I to BC and so is the point of tangency of BC with the incircle (Figure 8). Figure 8: 2 Then jA1P j = jA1A2jjA1A3j Proof We have jA1P j = jA1Bj ¡ jBP j a = ¡ (s ¡ b) (step 3) 2 b ¡ c = 2 b2 ¡ c2 jA A j = (step 5) 1 2 2a jA1A3j = jBA1j ¡ jBA3j a ac = ¡ (step 4) 2 b + c a(b ¡ c) = : 2(b + c) It follows that b ¡ c jA P j2 = jA A jjA A j = ( )2; 1 1 3 1 2 2 as required. We now return to the proof of Feuerbach's theorem which states that the nine point circle C9 of a triangle ABC is tangent to the incircle and the three escribed circles of the triangle. 7 Let A1 be the midpoint of the side BC and let P and Pa be the points of tangency of the incircle and the escribed circle drawn external to side BC; respec- tively (Figure 9). We consider the inversion mapping 2 2 2 Inv(A1; k ) where k = jA1P j and we denote it by Inv: Since P is the point of tangency of the side BC with the incircle C(I; r); then 2 jA1P j = ½C(I;r)(A1) By step 6, it follows that Inv(C(I; r)) = C(I; r) Since Pa is the point of tangency of the side BC with Figure 9: the escribed circle C(Ia; ra) and jA1Paj = jA1P j; then 2 2 2 ½C(Ia;ra)(A1) = jA1Paj = jA1P j = k ; then Inv(C(Ia; ra)) = C(Ia; ra): Thus C(I; r) and C(Ia; ra) are both invariant under the mapping Inv: Now we consider the image of the nine-point circle under Inv: Since A1 belongs to the nine-point circle C9 and A1 is the pole of Inv; then Inv(C9) is a line d. But C9 is the circumcircle of the triangle with vertices the midpoints A1;B1 and C1 of the sides of the triangle ABC so the line d is antiparallel to the line B1C1 (step 1). Since B1C1kBC then d is antiparallel to the side BC: 2 We also have that jA1A2jjA1A3j = jA1P j (step 7) and since A2 belongs to C9 then d is a line which passes through A3; as Inv(A2) = A3: Now let B0C0 be the second common tangent of the two circles C(I; r) and C(Ia; ra): Since A3 is the bisec- tor of the angle at A, these common tangents intersect b c0 0 at A3 (Figure 10).