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#1 h Mar 8, 2008, 9:56 PM • 1 Y

Y by Adventure10

Please any help.Thank you.

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#2 h Mar 10, 2008, 10:15 AM • 2 Y

Y by Adventure10, Mango247

Let $(O)$ be the circumcircle of $ABCD$ and $KLMN$ be an arbitrary quadrilateral (not necessarily a parallelogram), such that $R \in KM, P \in LN.$ Project the circle $(O)$ to a circle $(O')$ and the line $PR$ to infinity. The cyclic quadrilateral $ABCD$ goes into a cyclic parallelogram (rectangle) $A'B'C'D'.$ The lines $RKM, PLN$ go into parallels $K'M' \parallel B'C' \parallel D'A'$ and $L'N' \parallel A'B' \parallel C'D'.$ Let $K'L', M'N'$ meet $C'D', A'B'$ at $U', V'$ and let $K'L', M'N'$ meet $A'C'$ at $X', X''.$ By Menelaus theorem for the $\triangle A'C'P'$ ( $P'$ is at infinity) cut by the transversals $K'U'X', M'V'X'',$

$\frac {\overline{K'P'}}{\overline{K'A'}} \cdot \frac {\overline{X'A'}}{\overline{X'C'}} \cdot \frac {\overline{U'C'}}{\overline{U'P'}} = 1\ \Longrightarrow\ \frac {\overline{X'C'}}{\overline{X'A'}} = \frac {\overline{K'P'}}{\overline{K'A' }} \cdot \frac {\overline{U'C'}}{\overline{U'P'}} = \frac {\overline{U'C'}}{\overline{K'A'}}$

$\frac {\overline{V'P'}}{\overline{V'A'}} \cdot \frac {\overline{X''A'}}{\overline{X''C'}} \cdot \frac {\overline{M'C'}}{\overline{M'P'}} = 1\ \Longrightarrow\ \frac {\overline{X''C'}}{\overline{X''A'}} = \frac {\overline{V'P'}}{\overline{V'A' }} \cdot \frac {\overline{M'C'}}{\overline{M'P'}} = \frac {\overline{M'C'}}{\overline{V'A'}}$

$\frac {\overline{X'C'}}{\overline{X'A'}} \cdot \frac {\overline{X''A'}}{\overline{X''C'}} = \frac {\overline{U'C'}}{\overline{K'A'}} \cdot \frac {\overline{V'A'}}{\overline{M'C'}} = \frac {\overline{U'C'}}{\overline{M'D'}} \cdot \frac {\overline{V'A'}}{\overline{K'B'}} = \frac {\overline{L'C'}}{\overline{L'B'}} \cdot \frac {\overline{N'A'}}{\overline{N'D'}} = 1$

This means that the points $X' \equiv X''$ are identical, $K'L', M'N'$ meet at $X' \in A'C'$ and similarly, $L'M', N'K'$ meet at $Y' \in B'D'.$ Therefore, the original lines $KL, MN$ meet at $X \in AC$ and $LM, NK$ meet at $Y \in BD.$ If $KLMN$ is a parallelogram, $KL \parallel MN \parallel AC,$ $LM \parallel NK \parallel BD$ and $X, Y$ are originally at infinity.

In the projected case, $X', Y'$ are finite points outside of the rectangle $A'B'C'D'.$ WLOG, assume that $X', Y'$ are on the rays $(A'C', (B'D'.$ A unique conic $\mathcal K$ is tangent to the 5 lines $B'C', D'A', A'C', B'D', X'Y'.$ By our WLOG assumption, the pentagon formed by these 5 lines is convex and this conic is therefore an ellipse. Let $M''$ be arbitrary point on the segment $C'D'$ and let $M''Y', M''X'$ meet the segments $B'C', D'A'$ at points $L'', N''.$ Consider the hexagon $L''C'X'Y'D'N''$ with its main diagonals $L''X', C'D', X'N''$ concurrent at $M''$ and its 5 sidelines $L''C' \equiv B'C', C'X' \equiv A'C', X'Y', Y'D' \equiv B'D', D'N'' \equiv D'A'$ tangent to the ellipse $\mathcal K.$ By Brianchon theorem, the remaining sideline $L''N''$ is also tangent to this ellipse. But the ellipse $\mathcal K$ has a unique tangent $L'N' \parallel A'B' \parallel C'D'$ intersecting the segments $B'C', D'A',$ corresponding to a unique position of $M'$ on the segment $C'D'.$ Consequently, the quadrilateral $K'L'M'N',$ such that $X' \equiv K'L' \cap M'N' \cap A'C',$ $Y' \equiv L'M' \cap N'K' \cap B'D'$ and $L'N' \parallel A'B' \parallel C'D',$ $K'M' \parallel B'C' \parallel D'A'$ is unique. As a result, the original parallelogram $KLMN,$ such that $KL \parallel MN \parallel AC,$ $LM \parallel NK \parallel BD$ and $P \equiv LN \cap AB \cap CD,$ $R \equiv KM \cap BC \cap DA$ is also unique.

Let the internal bisectors of the angle $\angle BPC \equiv \angle DPA$ meet $BC, DA$ at $L, N$ and let the internal bisector of the angle $\angle ARB \equiv \angle CRD$ meet $AB, CD$ at $K, M.$ Let the angle bisectors $PLN, RKM$ meet at a point $Q.$ With reference to the posted figure, consider the concave quadrilateral $APQR.$

$\angle PQR = 360^\circ - (360^\circ - \angle A) - \frac {\angle PBC}{2} - \frac {\angle CRD}{2} =$

$= \angle A - \frac {180^\circ - (\angle B + \angle C)}{2} - \frac {180^\circ - (\angle C + \angle D)}{2} =$

$= \angle A + \angle C + \frac {\angle B + \angle D}{2} - 180^\circ = 90^\circ.$

This means that the $\triangle KPM, \triangle LRN$ are both isosceles and $Q$ is the midpoint of both $KM, LN.$ The resulting quadrilateral $KLMN$ has perpendicular diagonals cutting each other at half, hence, it is a rhombus.

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