rhombus from intersections of the angle bisectors with sides of cyclic ABCD

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rhombus from intersections of the angle bisectors with sides of cyclic ABCD

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Source: Almaty Olympiad 2009 p2 , Kazakhstan

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#1 h Sep 28, 2018, 5:17 PM • 1 Y

Y by Adventure10

The extensions of the sides $AB$ and $CD$ of the inscribed quadrilateral $ABCD$ intersect at the point $P$ , and the extensions of the sides $BC$ and $AD$ at the point $Q$ . Prove that the intersection points of the bisectors of the angles $AQB$ and $BPC$ with the sides of the quadrilateral are vertices of a rhombus.

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#2 h Sep 29, 2018, 8:14 AM • 1 Y

Y by Adventure10

Let the intersection points of angle bisectors be $X$ on $AD$ , $Y$ on $DC$ , $Z$ on $BC$ and $W$ on $AB$ . By Menelaus' theorem for triangle $BPC$ and transversal $A-D-Q$ we get
$\frac{|BA|}{|AP|} \cdot \frac{|PD|}{|DC|} \cdot \frac{|CQ|}{|QB|}=1.$ By power of a point theorem and similar triangles we get two equalities:
$\frac{|BA|}{|DC|}=\frac{|QA|}{|QC|},$ $\frac{|PD|}{|AP|}=\frac{|PB|}{|PC|}.$ Substituting that into the Menelaus' equation that we derived earlier, we get
$\frac{|QA|}{|QC|} \cdot \frac{|PB|}{|PC|} \cdot \frac{|QC|}{|QB|}=1,$ so $|QC|'$ s cancel out and we can see that
$\frac{|PD|}{|PA|}=\frac{|PB|}{|PC|}=\frac{|QB|}{|QA|}=\frac{|QD|}{|QC|}.$ From the angle bisector theorem we derive
$\frac{|AW|}{|WB|}=\frac{|QA|}{|QB|}=\frac{|PA|}{|PD|}=\frac{|AX|}{|XD|},$ so that means $XW$ and $DB$ are parallel. With similar analogy we can derive that $DB$ and $ZY$ are parallel, so that means that $XW$ and $ZY$ are parallel. Analogously follows $XY||AC$ , $AC||ZW$ , so $XY||ZW$ . Now it is clear that $XYZW$ is a parallelogram. Let $E$ be the intersection of its' diagonals. We know that $|WE|=|EY|$ , and since $E$ lies on angle bisector of $\angle {YPW}$ , we conclude that the triangle $WPY$ is isosceles and $PE$ is perpendicular to $WY$ . That means that $|WE|=|EY|$ and $ZE$ is perpendicular to $WY$ , so the triangle $WZY$ is isosceles and that means that all the sides of the parallelogram are equal. The problem is solved.

This post has been edited 1 time. Last edited by JANMATH111, Sep 29, 2018, 8:16 AM

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