Concyclicity in Rectangle
What is this about?
A Mathematical Droodle
Explanation
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Copyright © 1996-2012 Alexander Bogomolny
The applet attempts to suggest a problem by V. Proizvolov that I picked up in an issue of Quantum (September/October 1994). The magazine routinely carries three kinds of problems: math, physics, and brainteasers. The problme at hand, albeit mathematical to all extents and purposes, was included in the Brainteasers section (#B121). Perhaps it was too easy by the Quantum's standards. So here's the problem:
In rectangle ABCD, X is the midpoint of BC and Y the midpoint of CD; P is the intersection of DX and BY. Prove that ∠XAY = ∠DPY.
Proof
The problem is not difficult and is solved by "angle chasing".
∠DAY + ∠XAY + ∠BAX = 90°,
∠CDX = ∠BAX,
∠AYD = ∠BYC,
∠DAY + ∠AYD = 90°,
∠BAX + ∠AXB = 90°,
∠XPY = 360° - 90° - ∠BYC - ∠CXD, hence,
∠XPY = 90° + ∠DAY + ∠BAX, so that
∠DPY = 90° - ∠DAY - ∠BAX, but the same holds for ∠XAY,∴
∠XAY = ∠DPY, as required.
The problem is equivalent to saying that points A, X, P, Y are concyclic.
|Activities|
|Contact|
|Front page|
|Contents|
|Geometry|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
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