Jacobi’s Theorem (A Huge Trig Bash)

So here is Jacobi’s Theorem in all it’s glory: Let ABC be a triangle and let X, Y, Z be points in its plane such that \angle ZAB = \angle YAC, \angle ZBA = \angle XBC, and \angle XCB = \angle YCA. Then the lines AX, BY, CZ are concurrent.

Jacobi’s Theorem is pretty interesting because it trivializes the existence of points such as the First Napoleon Point, First Fermat Point, and in general, the Kiepert points. They are actually more simple cases of Jacobi’s Theorem because the triangles erected are all isosceles.

Proof:
Like the title suggests, the proof is a huge trig bash. For the sake of clarity, let \angle ZAB = \angle YAC = \alpha, \angle ZBA = \angle XBC = \beta, \angle XCB = \angle YCA = \gamma. We know that \dfrac{AZ}{\sin \angle ACZ} = \dfrac{CZ}{\sin \angle ZAC}. Rearranging, we obtain \sin \angle ACZ = \dfrac{AZ \sin\left(\angle CAB + \alpha\right)}{CW}. Similarly, \sin \angle ZCB = \dfrac{BZ \sin\left(\angle ABC + \beta\right)}{CZ}. Thus \dfrac{\sin \angle ACZ}{\sin \angle ZCB} = \dfrac{AZ \sin\left(\angle CAB + \alpha\right)}{BZ \sin\left(\angle ABC + \beta\right)}. We now apply Law of Sines again on \triangle ABZ to compute \dfrac{AZ}{BZ} = \dfrac{\sin \beta}{\sin \alpha}. Thus \dfrac{\sin \angle ACZ}{\sin \angle ZCB} = \dfrac{\sin\left(\beta\right) \sin\left(\angle CAB + \alpha\right)}{\sin\left(\alpha\right) \sin\left(\angle ABC + \beta\right)}. We are now done by Trig Ceva.

Alternate Proof:
Use the fact that AX, BX, CX concur at X. We obtain \dfrac{\sin \angle ABX}{\sin \angle XBC} \cdot \dfrac{\sin \angle BCX}{\sin \angle XCA} \cdot \dfrac{\sin \angle CAX}{\sin \angle XAB} = 1. We also have similar equations for Y and Z. Multiply them together and use the fact that the external segments are isogonal to get equal angles and you’re done.

Problem: Prove Kariya’s Theorem, which states that if D, E, F are the tangency points of the incircle of triangle ABC with the sides BC, CA, AB and if points X, Y, Z lie on lines ID, IE, IF with I being the incenter of ABC satisfying IX = IY = IZ and X, Y, Z all lie towards the interior or exterior of the triangle, then the lines AX, BY, CZ are concurrent.

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