Given triangle ABC and line ℓ passing through point A, point X on ℓ is considered to be variable. Circles ωb and ωc pass through both of the points A and X and are tangent to sides AB and AC, respectively. Tangents BY and CZ are drawn from vertices B and C to circles ωb and ωc respectively. Prove that as X varies, the circumcircle of triangle ZXY would be passing through two fixed points.
Solution
Let D be intersection of ℓ and BC. Let P and Q be the reflections of A with respect to BC and D, respectively. We claim that the circumcircle of XYZ passes through P and Q. Since PQ∥BC, we have ∠XQP=∠XDB, in line with BX=BY=BP, we have: ∠PYA⟹∠XYP=21∠PBA=∠ABC=∠AYP−∠AYX=∠ABC−∠DAB=∠ADB=∠XQP So Y lies on the circumcircle of XPQ. Similarly, Z also lies on this circle, and so we are done.
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