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.

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