Rhombus in Bicentric quadrilateral, please see problem in file ggb at here:

http://www.geogebratube.org/student/m79325

Solution by: Luis González:

Let $P \equiv BC \cap DA$ and $Q \equiv AB \cap CD.$ By angle chase, we have

$\widehat{IQP}+\widehat{IPQ}=\tfrac{1}{2}(\widehat{AQD}+\widehat{CPD})+\widehat{PQB}+\widehat{QPB}=$

$=\tfrac{1}{2}(\widehat{AQD}+\widehat{CPD})+\widehat{PBA}=180^{\circ}-\tfrac{1}{2}(\widehat{BCD}+\widehat{BAD})-\widehat{CDA}+...$

$=180^{\circ}-\tfrac{1}{2}(\widehat{BCD}+\widehat{BAD})=180^{\circ}-90^{\circ}=90^{\circ} \Longrightarrow \widehat{PIQ}=90^{\c...$

If $QI$ cuts $BC,DA$ at $F',H',$ then from the isosceles $\triangle PF'H',$ we get $\widehat{DH'F'}=\widehat{CF'H'}=90^{\circ}+\tfrac{1}{2}\widehat{CPD}=\widehat{DIC}$ $\Longrightarrow$ $H \equiv H'$ and $F \equiv F'.$ Analogously, $E,G$ are the intersections of $PI$ with $AB,CD.$ Hence $I$ is midpoint of $\overline{FH},\overline{EG}$ and $\widehat{EIF} \equiv \widehat{PIQ}=90^{\circ}$ $\Longrightarrow$ $EFGH$ is a rhombus.

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Thứ Bảy, 13 tháng 9, 2014

44-Rhombus in Bicentric

Rhombus in Bicentric quadrilateral, please see problem in file ggb at here:

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