A New Proof that the Spieker Point is the Radical Center of the Excircles of any Triangle.

    Theorem 1. The radical center of the 3 excircles of DABC is the Spieker point of DABC, namely, the incenter I' of DA'B'C'.
    Proof. Let the lines C'I' and AC intersect at S. We shall prove C' and S are midpoints of common tangents of K_A and K_B. First let Z_A and Z_B be the feet of perpendiculars from I_A and I_B to line AB. Then AZ_B = BZ_A = s - c, hence C' is the midpoint of Z_AZ_B. Next, let C'I'S cut A'B' at V, and let Y_A, Y_B be feet of perpendiculars from I_A and I_B to line AC. Then SB': B'A = B'V : AC'-B'V' = B'V : B'A'-B'V = B'V: A'V = B'C': A'C' = a : b. Hence SB' = b/2 × a/b = a / 2. Then SC = (a-b) / 2. From CY_B = s - a and CY_A = s - b we can deduce S is the midpoint of Y_AY_B.
    Since C' and S are midpoints of common tangents of K_A and K_B, C'I' is the radical axis of K_A and K_B. Similarly B'I' is the radical axis of K_A and K_C. Hence I' is the radical center of the 3 excircles. QED

    Theorem 2. Let B'A' be extended to P and C'A' be extended to Q so that A'P = A'Q = a / 2. Then P, Q lie on the circle centered at I' orthogonal to the 3 ex-circles of DABC . Also Z_APQY_A is a straight line.
    Proof. Let DABC have angles 2 a, 2 b, 2 g at A, B, C. Let N be the foot of perpendicular from I' to A'B'. Then I'P² = PN² + I'N² = (a/2 + (s-a)/2)² + (r/2 )² =(s² + r²)/4. I'P = I'Q is also trivial. So P,Q lie on the circle K' centered at I' with radius Ö(s² + r²) / 2.
    Next, we need to prove I'I_A² -I'Q² = r_A2 to show that K' is orthogonal to K_A. Let C'S cut I_AC at T. Notice C'T is perpendicular to IAC [either because C'I'TS is the radical axis and so is perpendicular to IAIB, or from computing angles]. Hence I'I_A² -I'Q² = I_AT² -QT² = (I_AQ+QT)² -QT² = I_AQ² + 2 I_AQ . QT .
    Now C'Q = b/2 + a/2. \QT = (½)(a + b) sin g . Again, since A'PQ is isosceles, PQ = 2 . (a/2) . sin a = a sin a. By angles chasing, we can see DPQI_A has angles g‘= p/2 -g, b‘=p/2 -b, a‘=p/2 -a respectively \I_AQ = PQ sin g‘ / sin a‘ = a tan a cos g. It follows that
    I_AQ² + 2 I_AQ . QT = a tan a cos g (a tan a cos g + 2 (½)(a + b) sin g ) . A most refreshing trigonometry ordeal will simplify this to s² tan²a = r_A².
    [Here are my proud details:
    a tan a cos g (a tan a cos g + (a + b) sin g )
= a tan a sec a cos g (a (sin a cos g + cos a sin g ) + b cos a sin g )
= a tan a sec a cos g (a cos b + b cos a sin g )
= a tan a sec a cos g ((4 R sin a cos a) cos b + (4 R sin b cos b) cos a sin g )
= a tan a cos b cos g 4 R ( sin a + sin b sin g )
= a tan a cos b cos g 4 R cos b cos g
= tan²a (4 R cos a cos b cos g)²
= s² tan²a .]
    To see that Z_APQY_A is a straight line, observe that C'Z_A = c/2 + (s - c) = (a + b)/2= C'Q, so that
ÐC'QZ_A = (p - 2a)/2 = ÐC'QP. Hence Z_APQ is a straight line. Similarly PQY_A is a straight line. QED

    Acknowledgment. Thanks to Paul Yiu of Florida Atlantic Univ. for directing me to these theorems.

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