Cherchez le quadrilatère cyclique
What is this about?
A Mathematical Droodle

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Explanation

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In a slightly different form the problem appeared in [Andreescu & Gelca, p. 9, problem 1.2.7]:

Points B and C are given on the side BC of a convex quadrilateral AEFD (with B closer to E than to F.) It is known that ∠BAC = ∠BDC and ∠BAE = ∠CDF. Prove that ∠CAF = ∠BDE.

The conclusion is valid even when points B and C coalesce whereas line BC is replaced by the tangent to the circle. With this in mind, the problem appears to generalize the one of an accidental angle bisector. Also, in this formulation, the requirement of convexity is quite spurious.

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Suffice it to show that ∠ADF + ∠AEF = 180°. This is so because

∠ADF + ∠AEF	= (∠ADC + ∠CDF) + (∠ABF - ∠BAE)
	= (∠ADC + ∠ABC) + (∠CDF - ∠BAE)
	= ∠ADC + ∠ABC
	= 180°.

References

1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004.

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

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