# Cyclic Quadrilateral, Concurrent Circles and Collinear Points

## What is this about?

A Mathematical Droodle

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Copyright © 1996-2017 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.

### Chasing Inscribed Angles

- Munching on Inscribed Angles
- More On Inscribed Angles
- Inscribed Angles
- Tangent and Secant
- Angles Inscribed in an Absent Circle
- A Line in Triangle Through the Circumcenter
- Angle Bisector in Parallelogram
- Phantom Circle and Recaptured Symmetry
- Cherchez le quadrilatere cyclique
- Cyclic Quadrilateral, Concurrent Circles and Collinear Points
- Parallel Lines in a Cyclic Quadrilateral
- Reim's Similar Coins I
- Reim's Similar Coins II
- Reim's Similar Coins III
- Reim's Similar Coins IV
- Pure Angle Chasing
- Pure Angle Chasing II
- Pure Angle Chasing III

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

Copyright © 1996-2017 Alexander Bogomolny

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