# Math circles and parallelograms

### Math circles and parallelograms

Given are two parallelograms ABCD and AECF with common
diagonal AC, where E and F lie inside the parallelogram ABCD.
Show:
The circumcircles of the triangles AEB, BFC, CED and DFA have a point
in common
Kokez

Posts: 7
Joined: Fri Jan 13, 2023 7:00 am
Reputation: 4

### Re: Math circles and parallelograms

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

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.
Guest

### Re: Math circles and parallelograms

could you please show me the drwan sketch

Kokez

Posts: 7
Joined: Fri Jan 13, 2023 7:00 am
Reputation: 4