I have found this simple method for construction of six concyclic points on sidelines of a triangle on April 6, 2006 and posted in Hyacinthos Geometry Forum.

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Detail as following:

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**Theorem:**

*If three circles concur at one Cevian point and their other three common points lie on three Cevian lines and these three circles cut triangle sidelines at six points then these six points are concyclic.*

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**Proof:**

Suppose **P** is a point on plane of triangle **ABC**; **(O _{a})**,

**(O**,

_{b})**(O**are three circles concur at

_{c})**P**and has three other common points:

**A**on Cevian

_{o}**AP**,

**B**on Cevian

_{o}**BP**,

**C**on Cevian

_{o}**CP**. It means:

**(O**is circumcircle of

_{a})**PB**,

_{o}C_{o}**(O**is circumcircle of

_{b})**PC**,

_{o}A_{o}**(O**is circumcircle of

_{c})**PA**.Suppose

_{o}B_{o}**(O**,

_{a})**(O**,

_{b})**(O**cut triangle sidelines

_{c})**BC**,

**CA**,

**AB**at

**A**,

_{1}**A**,

_{2}**B**,

_{1}**B**,

_{2}**C**,

_{1}**C**respectively. We proof these six points are concyclic.

_{2}.

We use power of circle theorem:

.

– Power of **A** wrt circles **(****O _{b}**

**)**and

**(O**:

_{c})**AC**

_{1}*****

**AC**=

_{2}**AA**

_{o}*****

**AP**=

**AB**

_{1}*****

**AB**therefore

_{2}**B**,

_{1}**B**,

_{2}**C**,

_{1}**C**are concyclic on one circle, suppose it is circle

_{2}**(K)**and

**(K)**cuts

**BC**at

**D**and

**E**.

.

– Power of **B** wrt circle **(K)**: **BC _{1}***

**BC**=

_{2}**BD***

**BE**

– Power of

**B**wrt circles

**(O**:

_{c})**BC***

_{1}**BC**=

_{2}**BB***

_{o}**BP**

Therefore

**BD***

**BE**=

**BB***

_{o}**BP**so

**D**,

**E**,

**B**,

_{o}**P**are on circle say

**(K**

_{1}).

– Power of **C** wrt circle **(K)**: **CB _{1}***

**CB**=

_{2}**CD***

**CE**

– Power of

**C**wrt circles

**(**

**O**

_{b}**)**:

**CB***

_{1}**CB**=

_{2}**CC***

_{o}**CP**

Therefore

**CD***

**CE**=

**CC***

_{o}**CP**so

**D**,

**E**,

**C**,

_{o}**P**are on circle say

**(**

**K**

_{2}**)**

.

Two circles **(K _{1})**,

**(**

**K**

_{2}**)**have three common points

**D**,

**E**,

**P**therefore they are the same one, say

**(K’)**. This circle

**(K’)**passes

**P**,

**B**,

_{o}**C**so it is the same circle

_{o}**(O**=

_{a})**(K’)**and

**D**,

**E**are intersections of

**(O**and sideline

_{a})**BC**. It means

**D**,

**E**are

**A**,

_{1}**A**and six points

_{2}**A**,

_{1}**A**,

_{2}**B**,

_{1}**B**,

_{2}**C**,

_{1}**C**are concyclic on one circle

_{2}**(K)**.

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**Remarks:**

1. Generally three circles **(O _{a})**,

**(**

**O**

_{b}**)**,

**(O**may be do not cut sidelines of

_{c})**ABC**but in each special case of

**P**we can choose proper circles

**(O**,

_{a})**(**

**O**

_{b}**)**,

**(O**such that they always cut sidelines of

_{c})**ABC**. For example:

2. If **P** is orthocenter **H**, we can choose **(O _{a})** as circle with center at midpoint of

**BC**and pass through

**H**.

**(**

**O**

_{b}**)**,

**(O**are similarly defined.

_{c})3. If

**P**is incenter

**I**, we can choose

**(O**as circle with center at touch point on

_{a})**BC**and pass through incenter

**I**.

**(O**,

_{b})**(O**are similarly defined.

_{c})4. With each **P**, we can choose some variants of **(O _{a})**,

**(**

**O**

_{b}**)**,

**(O**and will get some different interesting results.

_{c})