# Concyclicity in Rectangle

## What is this about?

A Mathematical Droodle

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Copyright © 1996-2018 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

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### 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|

Copyright © 1996-2018 Alexander Bogomolny

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