AoPS - Problem

Source:

Tags: geometry, angle bisector, Concyclic

parmenides51

26.09.2024 17:49

In a convex quadrilateral $ABCD$ we draw the bisectors $\ell_a$, $\ell_b$, $\ell_c$, $\ell_d$ of external angles $A$, $B$, $C$, $D$ respectively. The intersection points of the lines $\ell_a$ and $\ell_b$, $\ell_b$ and $\ell_c$, $\ell_c$ and $\ell_d$, $\ell_d$ and $\ell_a$ are designated by $K$, $L$, $M$, $N$. It is known that $3$ perpendiculars drawn from $K$ on $AB$, from $L$ om $BC$, from $M$ on $CD$ intersect at one point. Prove that the quadrilateral $ABCD$ is cyclic.

Useful Lemma: If we have two lines with $A$ on one and $B$ on the other, there are two circles tangent to the lines and segment $AB$. If they touch $AB$ at $X$ and $Y$, then $AX=BY$. Proof Outline: If the lines are parallel, then $X=Y$ is the midpoint of $AB$. Otherwise, connect the two lines at $C$ such that $AX,BX,AY,BY$ are among $s-a,s-b$. Solution: Without loss of generality, let the perpendiculars from $K$, $L$, and $M$ intersect at $I'$, and suppose they meet the $AB,BC,CD$ at $X',Y',Z'$ respectively. There is a circle centered at $I$ touching $AB,BC,CD$ at $X,Y,Z$, respectively. By the lemma, the midpoint of $KK'$ is that of $AB$ and so on. Therefore, all perpendicular bisectors of $AB,BC,CD$ meet at the midpoint of $II'$, and the quadrilateral is cyclic.