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At first we note an important result: Lemma 1. If A and B are two points in the plane, and X and Y are two other points, and Z is the point on the line XY which divides the segment XY in the ratio XZ : ZY = p : q (with directed segments), then $\left[ AZB\right] =\frac{q\cdot \left[ AXB\right] +p\cdot \left[ AYB\right] }{q+p}$, where $\left[P_1P_2P_3\right]$ denotes the directed area of a triangle $P_1P_2P_3$. ("Directed" means that the area is taken positive if the triangle is orientated counterclockwise, and negative if it is orientated clockwise.) Proof. Let x, y, z be the directed distances from the points X, Y, Z to the line AB. ("Directed" means that e. g. the distance x is positive if the triangle AXB is orientated counterclockwise, and negative if it is orientated clockwise.) Then, since the area of a triangle equals $\frac12 \cdot$ sidelength $\cdot$ corresponding altitude, we have $\left[ AXB\right] =\frac{1}{2}\cdot AB\cdot x$; $\left[ AYB\right] =\frac{1}{2}\cdot AB\cdot y$; $\left[ AZB\right] =\frac{1}{2}\cdot AB\cdot z$. Thus, the areas [AXB], [AYB], [AZB] are proportional to the distances x, y, z, respectively. Hence, instead of proving $\left[ AZB\right] =\frac{q\cdot \left[ AXB\right] +p\cdot \left[ AYB\right] }{q+p}$, it is enough to show $z =\frac{q\cdot x +p\cdot y}{q+p}$. But this is clear if you consider the line AB as the x-axis of an Cartesian coordinate system (then x, y, z are the x-coordinates of the points X, Y, Z, respectively). Lemma 1 is proven. Another important and well-known lemma: Lemma 2. If ABC is a triangle with sidelengths a = BC, b = CA, c = AB and circumradius R, then the area of triangle ABC equals $\frac{abc}{4R}$. Now, in our cyclic quadrilateral ABCD, let a = AB, b = BC, c = CD, d = DA, e = AC, f = BD. Let R be the circumradius of the quadrilateral ABCD. And finally, call p : q = AE : EB = CF : FD. We will apply Lemma 1 but work with non-directed areas. In other words, in the following, the notation $\left[P_1P_2P_3\right]$ will mean the non-directed area of a triangle $P_1P_2P_3$. Of course, one can apply Lemma 1 for non-directed areas only if the points X, Y, Z all lie on one and the same side of the line AB; but this will be always guarranteed in the following. Applied to triangle ABD, whose sidelengths are AB = a, BD = f and DA = d, and whose circumradius is R, Lemma 2 yields $\left[ ABD\right] =\frac{afd}{4R}$. Since the point E lies on the line AB and we have AE : EB = p : q, Lemma 1 gives $\left[ AED\right] =\frac{q\cdot \left[ AAD\right] +p\cdot \left[ ABD\right] }{q+p}=\frac{q\cdot 0+p\cdot \left[ ABD\right] }{q+p}=\frac{p\cdot \left[ ABD \right] }{q+p}$ $=\frac{p}{q+p}\cdot \left[ ABD\right] =\frac{p}{q+p}\cdot \frac{afd}{4R}$. Similarly, $\left[ AFD\right] =\frac{q}{q+p}\cdot \frac{ced}{4R}$. Since the point P lies on the line EF and we have PE : PF = AB : CD = a : c, Lemma 1 yields $\left[ APD\right] =\frac{c\cdot \left[ AED\right] +a\cdot \left[ AFD\right] }{c+a}=\frac{c\cdot \frac{p}{q+p}\cdot \frac{afd}{4R}+a\cdot \frac{q}{q+p} \cdot \frac{ced}{4R}}{c+a}=\frac{\left( pf+qe\right) acd}{4R\left( p+q\right) \left( c+a\right) }$. Similarly, $\left[ BPC\right] =\frac{\left( pf+qe\right) acb}{4R\left( p+q\right) \left( c+a\right) }$. Thus, $\frac{\left[ APD\right] }{\left[ BPC\right] }=\frac{d}{b}$, what is independent from the choice of E and F indeed. Proof complete. Darij

If $BC$ and $DA$ are parallel, then it's trivial, because then $ABCD$ would be an isosceles trapezoid, so in fact, we would have $AE=CF$ and then it's easy to see that $d(P,BC)=d(P,DA)$. So let's now assume that $BC$ and $DA$ intersect at a point $G$. We have $AB: CD=AE: CF=EP: FP$. But since $ABCD$ is cyclic, the triangles $GAB$ and $GCD$ are similar with ratio $AB: CD$. But since $AE: CF$ is equal to the similtude ratio, the point $E$ in $GAB$ corresponds to the point $F$ in $GCD$. Thus, the triangles $GFC$ and $GEA$ are similar with ratio $AB: CD$ so we have $GE: GF=AB: CD$ and $\angle DEF = \angle CGE$ Thus, $GE: GF=EP: FP$ and hence, $GP$ is the angle bisector of $\angle AGB$ from which we conclude $d(P,AD)=d(P,BC)$, yielding that the areas in question have a same attitude, thus their ratio is equal to the ratio of the corresponding sides, i.e. $AD: BC$ $\Box$

1998 G2 wrote: Let $ABCD$ be a cyclic quadrilateral. Let $E$ and $F$ be variable points on the sides $AB$ and $CD$, respectively, such that $AE:EB=CF:FD$. Let $P$ be the point on the segment $EF$ such that $PE:PF=AB:CD$. Prove that the ratio between the areas of triangles $APD$ and $BPC$ does not depend on the choice of $E$ and $F$. Tutorial for linearity Animate $E$ linearly on side $AB$. Then $E \mapsto F$ is linear and since $P$ divides $EF$ in a fixed ratio, $P$ moves linearly. Thus, it suffices to show that the locus of $P$ passes through $Q=AD \cap BC$, to conclude that $\frac{[APD]}{[BPC]}$ is fixed. Let $X$ and $Y$ be points on $AC$ and $BD$ such that $\frac{AX}{XC}=\frac{BY}{YD}=\frac{AB}{CD}$. When $E=A$ we have $P=X$ and when $E=B$ we have $P=Y$. Thus, $\overline{XY}$ is the locus of $P$. We want to show that $Q, X, Y$ are collinear. Notice that $\frac{QA}{QC}=\frac{QB}{QD}=\frac{AB}{CD}$ hence $X, Y$ lie on the internal bisector of angle $BQD$, proving the result. $\blacksquare$

Let $AE=x,BE=ax,CF=y,DF=ay.$ It follows that $AB:CD=x:y$ so $PE:PF=x:y.$ Define $d(X,YZ)$ to the distance between point $X$ to line $YZ$ for any $X,Y,Z.$ We have $d(E,AD)=x\sin(\angle A),d(F,AD)=ay\sin (\angle C).$ Since $P$ is $\frac{x}{x+y}$ of the way from $E$ to $F$ we have \[d(P,AD)=\frac{yd(E,AD)+xd(F,AD)}{x+y}=\frac{xy}{x+y}\left(\sin(\angle A)+a\sin\angle C\right)\]Similarly, we have \[d(P,BC)=\frac{xy}{x+y}\left(\sin(\angle C)+a\sin\angle D\right)\]and since $ABCD$ is cyclic, we see that these two are actually equal. Thus, we have \[\frac{[APD]}{[BPC]}=\frac{AD\cdot d(P,AD)}{BC\cdot d(P,BC)}=\frac{AD}{BC}\]which is not dependant on the choice of $a$ as desired.

G2? Let $\frac{AE}{AB}=\frac{CF}{CD}=k$. Note that \[[APD]=[AED]\cdot \frac{CD}{AB+CD}+[AFD]\cdot \frac{AB}{AB+CD}\]which simplifies to \[[APD]=k\cdot [ABD]\cdot \frac{CD}{AB+CD}+(1-k)\cdot [ACD]\cdot \frac{AB}{AB+CD}.\]Repeating similar calculations, we find that \[\frac{[APD]}{[BPC]}=\frac{k\cdot [ABD]\cdot CD+(1-k)\cdot [ACD]\cdot AB}{(1-k)\cdot [ABC]\cdot CD+k\cdot [BDC]\cdot AB}.\]It is then enough to show that \[\frac{[ABD]\cdot CD}{[BDC]\cdot AB}=\frac{[ACD]\cdot AB}{[ABC]\cdot CD}.\]But now we have $\frac{[ABD]\cdot CD}{[BDC]\cdot AB}=\frac{AB\cdot CD\cdot AD}{AB\cdot CD\cdot BC}=\frac{AD}{BC}$. The RHS is also equal to $\frac{AD}{BC}$, so we are done. $\blacksquare$

MMP > Note that as $E$ moves along $AB$, $F$ moves linearly along $BC$. Furthemore, $P$ is a weighted mean of $E$ and $F$, so it also moves linearly as $E$ moves. Let $P_1$ be the point on $AC$ such that $$\frac{AP_1}{P_1C}=\frac{AB}{CD}.$$Similarly, let $P_2$ be the point on $BD$ such that $$\frac{BP_2}{P_2D}=\frac{AB}{CD}.$$Finally, let $P_3$ be the intersection of $AD$ and $BC$ (possibly a point at infinity). If $E=A,$ then $F=C$ and $P=P_1$. Similarly, if $E=B$, then $F=D$ and $P=P_2$. Therefore, since $P$ moves linearly, $P$ always lies on line $P_1P_2$. We want to show that the ratio of the heights from $P$ to $AD$ and $BC$ is constant. Thus, it suffices to show that $P_1,P_2,P_3$ are collinear. Note that from similar triangles $\triangle P_3AB$ and $\triangle P_3CD,$ $$\frac{AB}{CD}=\frac{P_3A}{P_3C}.$$Hence, $$\frac{AP_1}{P_1C}=\frac{AB}{CD}=\frac{P_3A}{P_3C},$$so $P_1$ lies on the bisector of $\angle AP_3C$. Similarly, $P_2$ lies on the bisector of $\angle BP_3D$, so $P_1,P_2,P_3$ are collinear and we are done.

Let $d(X,l)$ denote the distance from point $X$ to line $l$. We desire to show that $d(P, AD) : d(P, BC)$ is constant regardless of $E,F$. We calculate $d(P,AD) =\frac{CD \cdot d(E, AD)+ AB \cdot d(F,AD)}{AB+ CD} = \frac{x \cdot CD \cdot AB \sin A + (1 - x) \cdot AB \cdot CD \sin B}{AB + CD} = \frac{AB \cdot CD (x \sin A + (1 - x) \sin B)}{AB + CD}$, $d(P, BC) = \frac{CD \cdot d(E, BC) + AB \cdot d(F,BC)}{AB + CD} = \frac{(1 - x) \cdot CD \cdot AB \sin B + x \cdot AB \cdot \sin A}{AB + CD} = \frac{AB \cdot CD (x \sin A + (1 -x) \sin B)}{AB + CD}$ as desired.