Let $ ABCD $ be a convex quadrilateral with no pair of parallel sides, such that $ \angle ABC = \angle CDA $. Assume that the intersections of the pairs of neighbouring angle bisectors of $ ABCD $ form a convex quadrilateral $ EFGH $. Let $ K $ be the intersection of the diagonals of $ EFGH$. Prove that the lines $ AB $ and $ CD $ intersect on the circumcircle of the triangle $ BKD $.
Problem
Source: Middle European Mathematical Olympiad 2012 - Team Compt. T-6
Tags: geometry, circumcircle, trapezoid, incenter, angle bisector, geometry proposed
WakeUp
14.09.2012 21:00
Let $E$ be the intersection of the bisectors of $\angle C$ and $\angle D$; let $F$ be the intersection of the bisectors of $\angle B$ and $\angle C$ (as in diagram). Now $\angle ABC=\angle CDA\implies\angle ADH=\angle ABG\implies$ $\triangle ADH\sim\triangle ABG\implies\angle EHG=\angle FGH$. Similarly $\triangle CED\sim\triangle CFB\implies\angle HEF=\angle HFG$. It follows that $EFGH$ is an isosceles trapezium with $EF||GH$. Let $EH$ and $FG$ meet at $X$. Then $XK$ is the bisector of $\angle FXE$ and so $\angle HXK=\angle KXF\implies\angle DXK=\angle BXK$. Now angle chasing shows that $\angle LDX=180^{\circ}-\frac{1}{2}D=180^{\circ}-\frac{1}{2}B=\angle XBL$ so $BDLX$ is cyclic. Note that $F$ must be the incentre of $\triangle BLC$ since $F$ lies on the angle bisectors of $\angle LBC$ and $\angle LCB$. So $LF$ is the bisector of $\angle BLC$ thus $\angle FLB=90^{\circ}-\frac{1}{2}B-\frac{1}{2}C$. By angle chasing, $\angle LFB=90^{\circ}+\frac{C}{2}$. Since $\angle EFX=\frac{1}{2}B+\frac{1}{2}C$, this means $\angle EFK=90^{\circ}-\frac{B}{2}\implies\angle FHG=90^{\circ}-\frac{1}{2}B$. Let $FH$ meet $AB$ at $L'$. Then $\angle HAL'=180^{\circ}-\frac{1}{2}A$ and $\angle AHL'=90^{\circ}-\frac{B}{2}$. Thus $\angle HL'A=\frac{1}{2}A+\frac{1}{2}B-90^{\circ}=90^{\circ}-\frac{1}{2}B-\frac{1}{2}C=\angle ALF$ (seeing as $A+C=360^{\circ}-2B$) so $\angle ALF=\angle AL'F$ which forces $L=L'$. Therefore $L$ lies on the same line as $H,F$ and $K$. This means that $KL$ is the angle bisector of $\angle BLD$ and $XK$ is the angle bisector of $\angle DXB$. Since $BDLX$ is cyclic, $K$ must be the midpoint of the arc $BC$ on this circle. Therefore $BKDL$ is cyclic.
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Matematika
17.09.2012 17:22
This problem is my own (Matija Bucić, Croatia) I am very interested in what you think about it (ok except that it is really easy I know that already but this competition in the end is made for second teams and younger students).
RaleD
17.09.2012 20:35
Let $DA$ intersects $CB$ at $Y$ and $L$ point where $AB$ intersects $CD$ . We have that $L, H, F$ are collinear and same holds for $Y, E, G$. ($HF$, $EG$ are angle bisectors of $\angle{DLA} $ and $ \angle{AYB}$) Also, we have that $Y, L, D, B$ are concyclic. We know that bisectors of $\angle{DLA} $ and $ \angle{AYB}$ intersect each other at circumcircle of $L, D, B, X$. Now it is obvious that $L, D, K, B$ are on one circle. Congrats Matija, I really liked this problem.
Kimchiks926
25.03.2022 17:53
Does this work?
Define points $X = AB \cap CD$ and $Y = AD \cap BC$. Obviosly, $\angle XBY = \angle XDY$, which implies that quadrilateral $XBDY$ is cyclic.
It will be assumed that the point $E$ is intersection of angle bisectors of $\angle A$ and $\angle B$, the point $F$ is intersection of angle bisectors of $\angle B$ and $\angle C$, the point $G$ is intersection of angle bisectors of $\angle C$ and $\angle D$, the $H$ is intersection of angle bisectors of $\angle D$ and $\angle A$.
Note that $H$ is the incenter of $\triangle ADX$, therefore $XH$ is angle bisector of $\angle AXD$. Moreover, the poing $F$ is the $X$ - excenter of $\triangle XBC$ meaning that $XF$ is the angle bisector of $\angle BXC = \angle AXD$. Thus the points $X,H,K,F$ are collinear. Similarly, we can show that points $Y,G,K,E$ are collinear. As the result angle bisectors $\angle BXC$ and $\angle CYD$ intersect at the point $K$, which is also the midpoint of smaller arc $BD$ of $\odot(XBDY)$. Therefore $K \in \odot(XBDY)$ and the problem is solved.