Let ABC be a triangle. Let B' and C' denote respectively the reflection of B and C in the internal angle bisector of ∠ A, Show that the triangle ABC and AB'C' have the same in centre?What should be the approach in this question? I don't even know how to initiate.

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Let ABC be a triangle. Let B&#039; and C&#039; denote respectively the reflection of B and C in the internal angle bisector of   ∠  A, Show that the triangle ABC and AB&#039;C&#039; have the same in centre?What should be the approach in this question? I don&#039;t even know how to initiate.

Carly Yang · Accepted Answer

Hint: Since it is a reflection, the angle bisector at   B reflects to the angle bisector at       B    ′    , and similarly for the bisectors at   C and at       C    ′    . In either triangle, the center of the incircle is at the intersection of the three bisectors. So each triangle&#039;s incenter lies on the bisector at   A, and maybe the above about the other two bisectors in each triangle will lead to something to prove it.

Emilio Calhoun · Accepted Answer

In-center I is the point of concurrence. Since one bisector is fixed and other two bisectors are swapped by reflection about the first fixed line , position of I does not change.

Let ABC be a triangle. Let B' and C' denote respectively the reflection of B and C in the internal angle bisector of angle A, Show that the triangle ABC and AB'C' have the same in centre? What should be the approach in this question?

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