Isogonal Conjugate Theorem

    Theorem: Let P be any point inside any triangle ABC with I as incenter. A geometric object K' is a reflection of another object K in a line L means K and K' are mirror images of each other in L.
     Let the line AX' be the reflection of AP in AI, cutting BC at X',
     Let the line BY' be the reflection of BP in BI, cutting CA at Y'.
     Let the line CZ' be the reflection of CP in CI, cutting AB at Z'.
     Then the lines AX', BY', CZ' will concur at some point P', called the isogonal conjugate of P in the triangle.

    First Proof. Let the lines AP, BP, CP cut BC, CA, AB at X, Y, Z respectively. Let x = BX, y = CY. Our goal is to compute (a - y)/y. First, let q = ÐBAX.= ÐCAX'. Then from the law of sines on DABX we get
     x / sin q = c / sin (q+B), yielding cot q = - cot B + c /(x sin B).
     Similarly, cot q = - cot C + b /(y sin C). Hence
- cot C + b / (y sin C) = - cot B + c / (x sin B),
     Using a / sin A = b / sin B = c / sin C to replace b and c, we get
sin A sin (C - B) / a + sin² B / y = sin '2 C / x,
which, after replacing A by p - B - C, simplifies to
(a - y) / y = a/y - 1 = (a - x) sin² C / x sin² B. (Exercise)
    In other words, BX' / X'C = (sin² C / sin² B) CX / XB . . . (1).
Similarly, CY' / Y'A = (sin² A / sin² C) AY / YC . . . (2),
and AZ' / Z'B = (sin² B / sin² A) BZ / ZA . . . (3).
     Multiplying (1), (2), (3), we get
(AZ' / Z'B) (BX' / X'C) (CY' / Y'A) = (BZ / ZA) (AY / YC) (CX / XB).
So by Ceva's theorem, AX', BY' CZ' are concurrent iff AX, BY, CZ are. QED

     We need two lemmas to prepare for the Second Proof.

    Lemma 1. Let A, B, P, Q be 4 distinct points on the plane. Then PQ ^ AB iff AP² - BP² = AQ² - BQ². [Proof. Necessity is trivial, but sufficiency takes some skill.]

    Lemma 2. Let ABC be any triangle.
Let L₁ be a line intersecting BC at right angles at D.
Let L₂ be a line intersecting CA at right angles at E.
Let L₃ be a line intersecting AB at right angles at F.
Then L₁, L₂, L₃ are concurrent iff
AF² - FB² + BD² - DC² + CE² - EA² = 0.
Proof. Le L₃ intersect L₂ at P. Then from Lemma 1,
AF² - FB² + BD² - DC² = AP² - PB² + BP² - PC² = AP² - PC², so that AP² - PC² = AE² - EC² iff P lies on L₁.

    Second Proof of Theorem. Let I be the incenter of ABC. Let D,E,F be the feet of perpendiculars from P to BC, CA, AB. Let AL be perpendicular from A to EF, BM be perpendicular from B to FD, and CN be perpendicular from C to DE. Observe that AEPF is concyclic, hence ÐFAP = ÐFEP = ÐEAL. \ line AL is the mirror image of line AP in AI. Hence
EL² - LF² = EA² - AF² . . . (1)
     Similarly we can prove
FM² - MD² = FB² - BD² . . . (2) and
DN² - NE² = DC² - CE² . . . (3).
     Adding (1), (2), (3) we get EL² - LF² + FM² - MD² + DN² - NE² = 0.
Hence AL, BM, CN are concurrent at some point P'. QED

    Reference. The Second Proof is inspired by my solution of Problem 2408 of Crux Mathematicorum 1(1999) proposed by Mansur Boase, student at St. Paul's School, London, UK.