# Antiparallel via Three Reflections

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A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

# Antiparallel via Three Reflections

The applet illustrates a problem from the *College Mathematical Journal* (**909**, by Francisco Javier García Capitán, Spain)

Let α, β, γ be the angular measures of angles BAC, ABC, ACB:

∠ALB = 180° - α/2 - β = γ + α/2.

Since D is the reflection of L in X (IX⊥BC), ∠IDL = γ + α/2. Also

∠IDB = 180° - (γ + α/2) = β + α.

Next, by the two reflections in BI and CI,

∠AB'I = ∠IDL = γ + α/2 and

∠AC'I = ∠IDB = β + α/2.

It follows that ∠AB'I + ∠AC'I =

From here,

∠IB'C' = ∠IC'B' = α/2.

Further,

∠AB'C' = ∠AB'I - ∠IB'C' = (γ + α/2) - α/2 = γ.

Similarly, ∠AC'B' = β which makes the line B'C' *antiparallel* to BC. As we know, this implies that quadrilateral BCC'B' is cyclic.

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

Copyright © 1996-2018 Alexander Bogomolny