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.
The two red marked angles are equal.
All the green marked angles are equal.
- 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.