Antiparallel via Three Reflections
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Copyright © 1996-2012 Alexander Bogomolny
Antiparallel via Three Reflections
The applet illustrates a problem from the College Mathematical Journal (909, by Francisco Javier García Capitán, Spain)
Let ABC be a triangle with incenter I. Let AI meet BC at L, and let X be the contact point of the incircle with the line BC. If D is the reflection of L on X, we construct B' and C' as the reflections of D with respect to lines BI and CI, respectively. Show that the quadrilateral BCC'B' is cyclic.
Let α, β, γ be the angular measures of angles BAC, ABC, ACB: α + β + γ = 180°. We chase the angles: in ΔABL,
∠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 = α + β + γ = 180°, making quadrilateral AB'IC' cyclic. From here, ∠B'IC' = 180° - alpha; and, since ΔB'IC' is isosceles,
∠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|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
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