All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Sixty is Special (Posted on 2017-08-18)
I is the incenter of triangle ABC with B' the intersection of the ray BI and side AC and C' the intersection of ray CI and side AB. Prove that AB'IC' is a cyclic quadrilateral if and only if angle BAC is 60 degrees.

 Submitted by Bractals No Rating Solution: (Hide) I first tried to prove that AB'IC' was cyclic if and only if the point I lies on the circumcircle of ΔAB'C', but that started to generate too much algebra. The following is more geometrical and the difficulty level drops from 4 to 2. Let α, β, and γ be the measures of ∠BAC, ∠CBA, and ∠ACB respectively. ∠AB'I and ∠AC'I are exterior angles of ΔB'CB and ΔC'BC respectively. Thus,    ∠AB'I = ∠B'CB + ∠CBB' = γ + β/2   and    ∠AC'I = ∠C'BC + ∠BCC' = β + γ/2   and Therefore,    α = 60°  <==>  β + γ = 120°                  <==>  3*(β + γ)/2 = 180°                  <==>  (γ + β/2) + (β + γ/2) = 180°                  <==>  ∠AB'I + ∠AC'I = 180°                  <==>  AB'IC' is cyclic. QED

 Subject Author Date Solution & a further result Jer 2017-08-27 17:58:50

 Search: Search body:
Forums (0)