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Let ABC be a triangle. Let B' and C' denotes respectively the reflection of B and C in the internal angle bisectors of angle A. How do I prove that the triangles ABC and AB'C' have same incenter.

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Hint. The incentre of AB'C' is obtained by reflecting the incentre of ABC about the bisector.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. - From Review $\endgroup$
    – A. A.
    Dec 10 '15 at 8:33
  • $\begingroup$ @Adolfo For pedagogical reasons, I felt in this instance that it would be more helpful to the OP to receive a hint, leaving room for them to discover some steps for themselves, than a complete answer. $\endgroup$
    – David
    Dec 10 '15 at 8:35
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Triangles $ABC$ and $AB'C'$ are symmetric over the angle bisector of $A$ and hence so do their incenters. Also since the incenters lies on the bisector they must be the same point.

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  • $\begingroup$ plz help me with picture. Actually I didn't understand question. $\endgroup$
    – mnulb
    Dec 10 '15 at 8:14

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