Cyclic Quadrilateral, Concurrent Circles and Collinear Points
What is this about?
A Mathematical Droodle
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Copyright © 1996-2015 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.)
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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 © 1996-2015 Alexander Bogomolny
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