Well, that's easy using polars. Let E be the point of intersection of the tangents to the circle at the points C and D, and let F be the point of intersection of the lines AC and BD. We then have to prove that $EF \perp AB$. Well, let O be the center of our circle, i. e. the midpoint of the diameter AB, and let P be the point where this diameter AB meets the chord CD. By a well-known property of polars, for any four points $A_1$, $A_2$, $A_3$, $A_4$ on a circle, the point of intersection of the lines $A_1A_2$ and $A_3A_4$ lies on the polar of the point of intersection of the lines $A_2A_3$ and $A_4A_1$ with respect to the circle. Applying this to our points A, C, D, B which lie on our circle, we see that the point of intersection of the lines AC and DB lies on the polar of the point of intersection of the lines CD and BA with respect to our circle. In other words, the point F lies on the polar of the point P with respect to our circle. The point E also lies on the polar of the point P, since the point E is actually (par definitionem) the pole of the line CD, and as the line CD passes through the point P, the pole E of the line CD lies on the polar of the point P. Thus, we see that both points F and E lie on the polar of the point P with respect to our circle. Hence, the line EF is the polar of point P with respect to our circle. By the definition of a polar, this yields $EF \perp OP$, or, in other words, $EF \perp AB$. Problem solved. Darij

orl wrote: Given four points $A,$ $B,$ $C,$ $D$ on a circle such that $AB$ is a diameter and $CD$ is not a diameter. Show that the line joining the point of intersection of the tangents to the circle at the points $C$ and $D$ with the point of intersection of the lines $AC$ and $BD$ is perpendicular to the line $AB.$ Let K=AC^BD and H=AD^BC. As H is the orthocenter of ABK, KH is orthogonal to AB. Let O be the point of intersection of the tangents to the circle at the points $C$ and $D$. It holds that <ODH = <ODB = <DAB = <DHK, as both complementary of the same angle <DBA. Since by similar arguments one can prove that <OCH = <CHK, we ha ve that O is the center of c(DHCK), i.e. O lies on HK, then we are done.

pestich wrote: We have another 'Polish joke' here. AB being a diameter makes points C and D feet of altitudes from intersection points. According to Archimedes Theorem all three of the altitudes will be concurrent at point A. Pestich. Does anybody understand this? Darij

No - any explanation ?

pestich wrote: AB being a diameter makes points C and D feet of altitudes from intersection points. According to Archimedes Theorem all three of the altitudes will be concurrent at point A. Pestich thought that we are interested in the intersection of AC and BD, where problem says the intersection of the tangents at C and D.

There's one thing that we have to be careful when working with this kind of problem - we are not given the order which the points A, B, C, D are in. As a result, different configurations must be considered. For instantance, sprmnt21's angles don't make very much sense on the diagram that I drew. Darij's method seems to avoid this problem. Or we could use directed angles if we were to approach it in a euclidean fashion.

billzhao wrote: For instantance, sprmnt21's angles don't make very much sense on the diagram that I drew. or, viceversa, the diagram you drew doesn't make very sense wrt the angles I referred in the proof . Of course I'm joking. I completely agree with you about the fact " that we have to be careful when working with this kind of problem". But in this case ... yes, as you remember, "we are not given the order which the points A, B, C, D are in" but it is quite usual, if not specified, consider the clockwise order or (also working in this case) the counterclockwise order. I agree with you that in an official contest is a must to consider all the "different configurations", but I, usually, don't give a complete solution to the problems. In most of cases, I try to propose an idea or, at the best, a sketch.

Ah well, let's stop doing banalities. Of course, most of the proofs using non-directed angles refer to one possible arrangement and work analogously for the other possible arrangements. Sprmnt's very nice solution of the Polish problem is one example of this rule. By the way, he could have completely avoided the need to distinguish different arrangements by using directed angles modulo 180 and thus tackling all possible arrangements at once. pestich wrote: Darij, Do not pretend not understanding me. I considered configuration with points C and D on opposite sides of AB. But proof is still the same for all possible arrangements of points C and D. All that's needed is the Archimedes theorem of altitudes concurrency in a triangle. Pestich. Pestich, I really don't see any "proof" in your post. I have asked you about the actual crux of your proof rather than about bagatelles like different arrangements. The problem asks for a proof of the fact that "the line joining the point of intersection of the tangents to the circle at the points C and D with the point of intersection of the lines AC and BD is perpendicular to the line AB". In your post, the point of intersection of the tangents to the circle at the points C and D isn't even mentioned. Darij

I know that the intersection point of the tangents at C and D was not mentioned. There is a good reason for it : it does not change the picture. In the orthocentric configuration this point will be a foot of a median or one of Euler points on altitude. Do you agree? Pestich.

pestich wrote: I know that the intersection point of the tangents at C and D was not mentioned. There is a good reason for it : it does not change the picture. In the orthocentric configuration this point will be a foot of a median or one of Euler points on altitude. Do you agree? Pestich. Yes. But you didn't mention it, and you didn't prove it (although the proof is quite simple). Anyway, once written down completely, this solution is exactly Sprmnt's solution. Darij

Let $O$ be the center of $\odot (ABCD)$, Let $AC \cap BD=Z$ and Let $AD \cap BC=Y$, let $DD \cap CC=X$, from Pascal's Theorem on $ACCBDD$, we have, $X-Y-Z$, Now, let $AB \cap CD =R$, $\implies$ $R \in AB$, from Brocard's Theorem, $YZ$ is the polar of $R$, but since, $X \in YZ$, hence, $XZ$ is also the polar of $R$ $\implies$ $XZ \perp AB$

Solution without polars Let $F’$ Be the intersection of $(AOC)$ and $(BOD)$. Observer that $F’$ is the image of F after inversion with respect to $(ABCD)$. So $O,F,F’$ are collinear and $OD^2=OC^2=OF.OF’ (*)$. By angle chasing, $F’,D,F,C$ are concyclic. From $(*)$, we get that $E$ is the circumcenter of $(CFD)$. Now we can finish by angle chasing.

It's just a Pascal on $ACCBDD$ (note that $AD\cap BC$ - orhocenter of $[AC\cap BD]AB$).

Let $P=CC \cap DD ; R=AD \cap BC ; Q=AC \cap BD ; T=AB \cap CD$ By Brocard theorem: $RQ \perp AB$ , $RQ \equiv t$ By polars: $DT \equiv p \implies P \equiv t$ $\implies P,Q,R$ are collinear in the polar of $T$. $\implies PQ \perp AB$.$\blacksquare$