# Cyclic Quadrilateral, Concurrent Circles and Collinear Points

## What is this about?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

The applet attempts to suggest the following problem:

In a cyclic quadrilateral ABCD side AB and CD (when extended) meet at point P; sides BC and AD, at Q. Prove that

- The circumcircles C(ABQ), C(BCP), C(ADP), C(CDQ) are concurrent.
- The point of concurrency, say K, is collinear with P and Q.

(The problem is not original, but I have misplaced my note on the source.)

The fact that the four circles concur is the subject of Miquel's Theorem. The circles are concurrent even if ABCD is not cyclic.

Since it is cyclic,

∠BCD + ∠BAD = 180°.

The angles at points A and C are supplementary:

∠BCD + ∠BCP = 180°, and

∠BAD + ∠BAQ = 180°.

In circle C(BCP),

∠BCP + ∠BKP = 180°.

In circle C(ABQ),

∠BAQ + ∠BKQ = 180°.

Adding all the above identities (with written right-to-left) gives,

∠BKP + ∠BKQ = 180°,

which exactly means that point P, K, Q are collinear.

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

Copyright © 1996-2018 Alexander Bogomolny