Let r be the radius of the circle. Obviously SinA=SinC, SinB=SinD, so the problem is reduced to AC SinB = BD Sin A. But AC=2rSinB, BD=2rSinA, so ACSinB= BDSinA=2rSinASinB

Yes, of course, but the converse is (in my opinion) substantially more difficult.

Thats's why it's sometimes easier to work backwards from what you are trying to prove. Sort of like starting a maze from the end.

After all that, I got it out myself. If $A+C > 180$, then $AC \sin A > BD \sin B$ and also $AC \sin A > BD \sin B$, so $AC \sin A = \frac{AC.BK \sin A}{BD} + \frac{AC.DK \sin A}{BD} > BK \sin B + DK \sin D$, and doing a similar thing with $C$ we easily get that $AK \sin A+CK \sin C > BK \sin B + DK \sin D$.

I think your assertion that if $A+C>180$, then $AC \sin A<BD \sin B$ could be wrong. Suppose $A$,$B$,$D$ are on circle $O$ with radius $r$, while $C$ is inside the circle, then $A+C>180$. The segment $BC$ is extended and crosses the circle at $E$. Now we have \[AC \sin A=\frac {AC \cdot BD}{2r}\],\[BD\sin B=\frac {AE \cdot BD}{2r}\]. Therefore if $AC>AE$(which is absolutely possible), then $AC \sin A>BD \sin B$. The conclusion turns to be in the opposite.

Fiachra wrote: ABCD is a convex quadrilateral. The diagonals AC and BD intersect at K. Show that ABCD is cyclic if and only if $AK \sin A + CK \sin C=BK \sin B + DK \sin D$. Obviously the "only if" part is trivial. Actually, the IMO Shortlist 1997 problem 23 asked to prove that if AK sin A + CK sin C = BK sin B + DK sin D, then the quadrilateral ABCD is cyclic. This is a really diabolic problem, since it is quite easy to find a solution that works for the "usual" case (in most solutions, this is the case when the triangles ABC, BCD, CDA and DAB all are acute-angled, i. e. when each diagonal of the quadrilateral ABCD and each side form an acute angle), but not all of these solutions properly work in the general case of an arbitrary convex quadrilateral ABCD. Anyway, you can look at http://www.mathlinks.ro/Forum/viewtopic.php?t=15584 post #2 file IMO_1997_shortlist.pdf to find the proposed solution of the problem (scroll down to page 52), and you can also look at http://www.kalva.demon.co.uk/short/soln/sh9723.html (that's practically the same solution). Both times, you will find solutions working for arbitrary convex quadrilaterals ABCD. Proving the converse, i. e. proving that if ABCD is a cyclic quadrilateral, then AK sin A + CK sin C = BK sin B + DK sin D, is really a piece of cake, and it was done by Zengmao in post #2 of this thread. It could also be done as follows: Let X, Y, Z, W be the orthogonal projections of the point K on the sides AB, BC, CD, DA of the quadrilateral ABCD. Then, since < AWK = 90° and < AXK = 90°, the points W and X lie on the circle with diameter AK. In other words, the circumcircle of triangle AWX has diameter AK. Thus, by the extended law of sines (which states that a sidelength of a triangle equals the diameter of the circumcircle multiplied with the sine of the opposite angle), applied to the triangle AWX, we have $WX=AK\cdot\sin\measuredangle WAX$. In other words, $WX=AK\cdot\sin A$. Similarly, $XY=BK\cdot\sin B$, $YZ=CK\cdot\sin C$ and $ZW=DK\cdot\sin D$. Hence, instead of proving the equation AK sin A + CK sin C = BK sin B + DK sin D, it will suffice to establish the equation WX + YZ = XY + ZW. Since the points W and X lie on the circle with diameter AK, the chordal angle theorem yields < KWX = < KAX. In other words, < KWX = < CAB. Similarly, < KWZ = < BDC. But since the quadrilateral ABCD is cyclic, the chordal angle theorem yields < CAB = < BDC. Thus, < KWX = < KWZ. It follows that the point K lies on the angle bisector of the angle ZWX. Similarly, the point K lies on the angle bisectors of the angles WXY, XYZ and YZW. Thus, the angle bisectors of the angles WXY, XYZ, YZW and ZWX of the quadrilateral XYZW concur at one point (namely, at the point K). Hence, the quadrilateral XYZW has an incircle (with center K); thus, WX + YZ = XY + ZW. And this completes the proof. I'm actually posting this proof because it gives a subtle hint towards the proof of the rather difficult part of the problem (AK sin A + CK sin C = BK sin B + DK sin D $\Rightarrow$ the quadrilateral ABCD is cyclic)... Darij

I have found that the problem 2149 of CRUX MATHEMATICORUM it is asked prove that a proposition equivalent to problem of Shortlist. How I can To attach files?

Where do you get problems from crux???

For my own ease in labeling, let $K=X$. Drop perpendiculars from $X$ to $AB, BC, CD, DA$ with feet $K, L, M, N$ respectively. Then the midpoint of $AX$ is the center of the circle that passes through points $A, K, X, N$. Also, $KN$ subtends an angle $2\angle A$ at the center, so $KN=2r\sin\frac{2A}{2}=XA\sin A$ where $r$ is the radius of the circle. From the given condition in the question we have $KN+LM=MN+KL (**)$. But we also have \begin{align*}KN&=XK\cos\angle XKN+XN\cos\angle XNK\\&=XK\cos\angle XAN+XN\cos\angle XAK (*)\end{align*} Now, let $\angle XAD=\alpha_1$, $\angle XAB=\alpha_2$, $\angle XBA=\beta_1$, $\angle XBC=\beta_2$, $\angle XCB=\gamma_1$, $\angle XCD=\gamma_2$, $\angle XDC=\delta_1$, $\angle XDA=\delta_2$. Then $(*)$ becomes $KN=XK\cos\alpha_1+XN\cos\alpha_2$. Similarly, $KL=XL\cos\beta_1+XK\cos\beta_2$, $LM=XM\cos\gamma_1+XL\cos\gamma_2$, $MN+XN\cos\delta_1+XM\cos\delta_2$. So $(**)$ becomes \[XK(\cos\alpha_1-\cos\beta_2)+XL(\cos\gamma_2-\cos\beta_1)+XM(\cos\gamma_1-\cos\delta_2)+XN(\cos\alpha_2-\cos\delta_1)=0 (***).\] WLOG, assume $\alpha_1\le\beta_2$ i.e. $A$ lies outside the circumcircle of $\triangle BCD$, which means $\alpha_2\le\delta_1$. But we have $\alpha_1+\delta_2=\beta_2+\gamma_1$, so $\gamma_1\le\delta_2$. Similarly $\alpha_2+\beta_2=\gamma_2+\delta_1$, so $\gamma_2\le\beta_1$. The cosine function decreases over $[0^{\circ}, 180^{\circ}]$, so it follows that all the terms within parenthesis in $(***)$ are nonnegative. Hence they must all be zero, and all pairs of angles are equal. It follows that $ABCD$ is cyclic, as desired. $\Box$

ABCD cyclic sinA=sinC sinB=sinD we need prove that ACsinA=BDsinB AC=2RsinB BD=2RsinA proved.