Concyclic Incenters in Bicentric Quadrilateral
What Might This Be About?
Problem
Given convex bicentric quadrilateral $ABCD,$ let $I$ and $O$ be its in- and circumcenter. Define $E,$ $F,$ $G,$ $H$ to be the incenters of triangles $OAB,$ $OBC,$ $OCD,$ $ODA,$ respectively.
Then $EFGH$ is a cyclic quadrilateral.
Solution
This is a special case of another problem where the quadrilateral $ABCD$ was just inscriptible. In that problem, the perpendicular bisectors formed an inscriptible quadrilateral whose role in the present problem is played by the circumcenter $O.$ This is naturally so because now the perpendicular bisectors of all four sides meet at $O$ - degenerate inscriptible quadrilateral.
Acknowledgment
The problem has been posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page.
Bicentric Quadrilateral
- Collinearity in Bicentric Quadrilaterals
- Easy Construction of Bicentric Quadrilateral
- Easy Construction of Bicentric Quadrilateral II
- Projective Collinearity in a Quadrilateral
- Fuss' Theorem
- Line IO in Bicentric Quadrilaterals
- Area of a Bicentric Quadrilateral
- Concyclic Incenters in Bicentric Quadrilateral
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