by Guest » Thu Feb 16, 2023 3:19 pm
Let O be the intersection point of the diagonals AC and BD. Since ABCD is a parallelogram, we have BD || AC, so ∠AOD = ∠BOC.
Now, consider the circumcircles of triangles AEB and CED. Let P be their intersection point (other than E and C, respectively). By the Inscribed Angle Theorem, we have:
∠AEP = ∠ABP
∠CEP = ∠CDP
Subtracting these two equations, we get:
∠AEP - ∠CEP = ∠ABP - ∠CDP
∠AEC = ∠ABD
Since AB || CE, we have ∠AEC = ∠ADE. Therefore, ∠ADE = ∠ABD, which means that the quadrilateral ABED is cyclic. Similarly, the quadrilateral BCFD is also cyclic.
Let Q be the intersection point of the circumcircles of triangles AEB and BFC (other than B). Then, by the Inscribed Angle Theorem, we have:
∠AEQ = ∠ABQ
∠BFQ = ∠BCQ
Adding these two equations, we get:
∠AEQ + ∠BFQ = ∠ABQ + ∠BCQ
∠AEF = 180° - ∠ACB
Since ∠AEF is constant (it does not depend on the position of E and F inside ABCD), this means that the circumcircles of triangles AEB and BFC always intersect at the same point Q, regardless of the position of E and F.
Similarly, let R be the intersection point of the circumcircles of triangles CED and DFA (other than D). Then, by the Inscribed Angle Theorem, we have:
∠CER = ∠CDR
∠AFR = ∠ADR
Adding these two equations, we get:
∠CER + ∠AFR = ∠CDR + ∠ADR
∠CFE = 180° - ∠ACB
Again, since ∠CFE is constant, this means that the circumcircles of triangles CED and DFA always intersect at the same point R, regardless of the position of E and F.
Therefore, we have shown that the circumcircles of triangles AEB, BFC, CED, and DFA have two common points, namely Q and R. However, these four circles cannot intersect at more than two points (by the Radical Axis Theorem), so Q and R must be the same point. Therefore, the circumcircles of triangles AEB, BFC, CED, and DFA have a point in common.