In 3 intersecting circles, choose the correct cyclic point which is also the intersection of two produced chords.

Question

Let ABXY and CDXY be two intersecting circles. It is also given that AYC is a straight line. AB produced will meet CD produced at V. It is easy to prove that VBXD is cyclic.

Conversely, suppose that ABXY, CDXY, and BXD are three intersecting circles. It is also given that AYC is a straight line. Let $V, V_1, V_2, ... V_n$ be some points on the circle BXD. Among these points, there is one and only one point happens to be the intersection point of the extended chords (AB and CD). The question is “How to prove that the point chosen is the right one?”.

In other words, if $V$ is the chosen one, then it is required to prove both $ABV$ and $CDV$ are straight lines.

Extra condition(s) are welcome to support what has been chosen. For example, we can always construct the fourth circle passing through ACV, and it will cut the circle VBXD at some point U. Will that help?

Answer 1

Assuming the question allows for construction, on $BX$ construct$$\angle XBV=\angle XYA=\angle XDC$$Since $\angle XDC$ is supplement to $\angle CYX=\angle ABX$, then $ABV$ is a straight line.

Now join $DV$. Since $\angle VDX$ is supplement to $\angle XBV=\angle XDC$, then $CDV$ is a straight line.

Therefore $V$ is the point on circle $BXD$ where $AB$ and $CD$ extended meet.

Answer 2

Construction:

(1) Extend YX [line 1], the radical axis of (ABXY) and (CDXY), to cut the circle BXD at W; (2) Through W, draw [line 2] parallel to AYC and let it cut the circle (BXD) at some point [V]. Claim: V is the required point. Firstly, we have to prove ABV is a straight line.

Proof:

(1) BXWV by itself is a cyclic quadrilateral; (2) $\gamma$ is an interior angle of that quadrilateral; (3) $\alpha$ is an angle “externally opposite” to $\gamma$; and (3) $\alpha$ happens to be equal to $\gamma$ via $\beta$. Thus, $\angle ABX$ is an angle exterior (and opposite) to the interior angle $\angle VWX$ and therefore ABV is a straight line.

Once the above is true, we are left with the following figure.

Proving CDV is straight becomes elementary.