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.

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  1. The two red marked angles are equal.

  2. All the green marked angles are equal.

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  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 $\alpha = \beta$.

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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.

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  • $\begingroup$ 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) $\endgroup$ – Oai Thanh Đào Jul 23 '16 at 17:17
  • $\begingroup$ @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. $\endgroup$ – Mick Jul 24 '16 at 4:20

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