# Two Circles and Two More

The applet below illustrates the following statement:

Two circles C(E) with center E and C(F) with center F intersect in points P and Q. Point A is on C(E), point B is on C(F) and AB passes through P. Let W be the intersection of AE and BF. Then E, F, Q, W are concyclic as are A, B, Q, W. In particular, W is the common point of the circumcircles

What if applet does not run? |

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Copyright © 1996-2018 Alexander Bogomolny

### Proof

First of all, observe that inscribed angles PAQ and PBQ do not depend on the position of AB as they are subtended by the fixed arcs PQ in C(E) and C(F), respectively. Therefore ∠AQB, the third angle in ΔAQB is also independent of the position of AB.

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Further, triangles AEP and BFP are isosceles,acute,equal,isosceles,equilateral with one of the base angles (PAQ in C(E) and PBQ in C(F)) independent of the position of AB. The same holds therefore for the other pair of the base angles, ∠APE and ∠BPF. It follows that their supplement at P, i.e., ∠EPF is also fixed:

∠EPF = 180° - ∠APE - ∠BPF. |

In quadrilateral EPFW, ∠PEW = 2∠APE and ∠PFW = 2∠BPF. So that,

∠EWF | = 360° - ∠PEW - ∠PFW - ∠EPF | |

= 180° - ∠APE - ∠BPF | ||

= ∠EPF, |

and so ∠EWF is independent of AB. Now ∠EWF is just the same as ∠AWB. We conclude that on one hand points A, Q, W, B are concyclic as are E, Q, W, F.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny