# Show that three circles are coaxal

Let $A_1, A_2, A_3, A_4$ are collinear, $B_1, B_2, B_3, B_4$ are collinear. Such that $A_1, A_2, B_2, B_1$ lie on circle $(O_1)$, and $A_3, A_4, B_4, B_3$ lie on circle $(O_2)$. Let $MNPQ$ be the quadrilateral form by $A_1B_1$, $A_2B_2$, $A_3B_3$, $A_4B_4$. Then show that $MNPQ$ is concyclic, and three circles $(MNPQ)$, $(O_1)$, $(O_2)$ are coaxal.

1. The two red marked angles are equal.

2. All the green marked angles are equal.

1. Then, the blue marked angles will be equal too. Therefore, MNPQ is con-cyclic.

The following is what I have tried for the second part:-

To every 2 circles (like $(O_1)$ and the red), there is always a radical axis. Let H be a point on this axis. Then HU (the tangent from H to the green circle at U) and also HV are equal in length. The question is:- “will HW = HU?”. This can be realized if $\alpha = \beta$.

After (1) producing $UO_1$ to cut $WO_2$ extended at K; (2) joining HK and (3) joining UW, we have HUKW being con-cyclic.

Construction: Using UK as diameter, construct the black circle.

By angle in the same segment, we have $\alpha = \alpha’$ and $\beta = \beta’$.

By angle in alternate segment, we should have $\alpha = \beta’$. The goal is then attained.

I am stuck at “will the black circle, HK and UW meet at the same point?” Any help is appreciated.

• When I try to fine a generalization of the Thebault theorem, I found this result. But, the result was found by Tran Quang Hung. (Tran Quang Hung found this some year ago) – Oai Thanh Đào Jul 23 '16 at 17:17
• @OaiThanhĐào Thank for the info. This is the first time I've heard of such theorem. How can I apply it to the present problem. At first glance, the three circles do not satisfy the requirement of that theorem. – Mick Jul 24 '16 at 4:20