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An Angle Bisection

​
∠AHB'
≈
47.55°
∠CHB'
≈
47.55°
Let ABC be a triangle and let A', B', and C' be the contact points of the incircle opposite A, B, and C, respectively. Let H be the foot of the perpendicular to A'C' from B'. Then B'H bisects
∠
AHC.
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