I know I once gave a pretty neat solution, but I definitely can't remember it now so I'll try to post the outline of a not-so-nice solution (it has no ingenious ideas ). (1) Try to find the locus of $K$, the midpoint of $CD$, as th line $CD$ varies. You'll see that it's the circumcircle of $OAB$, where $O$ is the midpt of $O_1O_2$ (the centers of the 2 circles). (2) Find the locus of $P$, the midpoint of $MN$, as $CD$ varies. It's the circle with diameter $O_1O_2$, so it also has center $O$. (3) Show that $PO\perp CD$ and we're done, because $O$ is on the perpndicular bisector of $AK$ (from (1)), and from (3) we find that the line $OP$ is the perpendicular bisector of $AK$, so $PA=PK$, so $K$ is on th circle centered at $P$ which passes through $A$, and that's precisely what we want, because $\angle MKN=90\Leftrightarrow K\in C(P,PA)$ (this is because $\angle MAN=90$).

O1O2 is center of two circle meet A,B O1M cut CB at P , O2N cut BD at Q then Then <NAM=90 then KP/NQ=BP/QK,then <PKM=<QNK => KNQ=90+<CPQ+<QDK=>MKN=90

A different solution: Let the mid-points of BC and BD be M' and N' respectively. Then M'KN'B is a parallelogram, $MM' \bot BC$ and $NN' \bot BD$. $\angle BNN'=\frac{1}{2} \angle BND=\frac{1}{2} \angle BAC=\angle MAC=\angle MBM'$. Thus the two right triangles BNN' and MBM' are similar. Then, $\frac{N'N}{M'K}=\frac{N'N}{BN'}=\frac{BM'}{M'M}=\frac{KN'}{M'M}$, giving $\frac{N'N}{KN'}=\frac{M'K}{MM'}$. In addition, $\angle NN'K=90^o +\angle DN'K =90 ^o +\angle CM'K =\angle KM'M$ Thus $\triangle NN'K \sim \triangle KM'M$. So, $\angle MKN=\angle MKM'+\angle M'KN'+\angle N'KN=\angle MKM'+\angle CM'K+\angle M'MK=180^o -\angle CM'M=90^o$

Although we already have 3 solutions for the original problem now, let me show a rather uncommon solution using inversion: Problem. Two circles c and d intersect each other at two points A and B, and a line l passing through the point A meets these two circles at the points C and D (apart from the point A). Let M and N be the midpoints of the arcs BC and BD which do not contain A on the circles c and d, respectively. Let K be the midpoint of the segment CD. Prove that < MKN = 90°. Solution. The crucial point of the solution will be an inversion with center A. But before we undertake this inversion, we transform the problem into a more convenient form: The point M is the midpoint of the arc BC on the circle c. Thus, the two chords BM and CM of the circle c have equal length. Hence, the chordal angles of these two chords must also be equal; in other words, we have < BAM = < CAM. Hence, the point M lies on the angle bisector of the angle BAC. Similarly, the point N lies on the angle bisector of the angle BAD. But, of course, the angle bisector of the angle BAD is the exterior angle bisector of the angle BAC; hence, since the interior and the exterior angle bisector of an angle are always perpendicular to each other, the angle bisector of the angle BAC and the angle bisector of the angle BAD are perpendicular to each other. Hence, < MAN = 90°. The problem requires to prove < MKN = 90°. Since we know that < MAN = 90°, this will trivially follow if we show < MKN = < MAN. But this is equivalent to proving that the points M, N, A and K lie on one circle. Hence, in order to solve the problem, it is enough to prove that the points M, N, A and K lie on one circle. Now, perform an inversion with respect to an arbitrary circle centered at the point A. Denote by B', C', D', M', N' and K' the images of the points B, C, D, M, N and K under this inversion. What properties do these new points B', C', D', M', N' and K' have? - We know that the circle c passes through the points A, B, C and M. Since this circle passes through the point A, which is the center of the inversion, it is mapped by the inversion to a line; this line must pass through the images B', C' and M' of the points B, C and M under our inversion. Thus, the points B', C' and M' lie on one line. Similarly, the points B', D' and N' lie on one line. - We know that the point M lies on the angle bisector of the angle BAC. Since our inversion has the center A, the images B', C' and M' of the points B, C and M under this inversion must lie on the rays AB, AC and AM, respectively, and thus we can state that the point M' lies on the angle bisector of the angle B'AC'. Similarly, the point N' lies on the angle bisector of the angle B'AD'. - The point K is the midpoint of the segment CD. This seems to be a property hard to deal with using inversion. However, there is a workaround: We denote by $\infty_{l}$ the infinite point of the line l (this is the point of intersection of the line l with the line at infinity). Then, the image of this point $\infty_{l}$ under our inversion is the point A (since an inversion maps infinite points to the center of the inversion). Also, since every segment is divided harmonically by its midpoint and the infinite point of the line containing the segment, the points K and $\infty_{l}$ divide the segment CD harmonically (since the point K is the midpoint of the segment CD, and the point $\infty_{l}$ is the infinite point of the line l = CD). In other words, the points C, D, K and $\infty_{l}$ on the line l form a harmonic quadruple. Now, it is well-known that if four points lying on a line through the center of an inversion form a harmonic quadruple, then so do their images under the inversion. Since the line l passes through the center A of our inversion, we thus conclude that the points C', D', K' and A, being the images of the points C, D, K and $\infty_{l}$ under our inversion, must also form a harmonic quadruple. Thus, the point K' is the harmonic conjugate of the point A with respect to the segment C'D'. - Finally, let's see how the inversion transforms the assertion that we have to prove. We have already seen that, in order to solve the problem, it suffices to show that the points M, N, A and K lie on one circle. In other words, it suffices to show that the points M, N and K lie on one circle through the point A. Since the point A is the center of our inversion, and since an inversion maps circles through the inversion center to lines and conversely, it is clear that proving that the points M, N and K lie on one circle through the point A is equivalent to showing that their inversive images M', N' and K' lie on one line. Hence, in order to solve the problem, it will be enough to show that the points M', N' and K' lie on one line. Now, everything is simple: Since the points B', C' and M' lie on one line, and the point M' lies on the angle bisector of the angle B'AC', the point M' is the point of intersection of the angle bisector of the angle B'AC' with the line B'C'. Since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we have $\frac{B^{\prime}M^{\prime}}{M^{\prime}C^{\prime}}=\frac{AB^{\prime}}{C^{\prime}A}$. Similarly, the point N' is the point of intersection of the angle bisector of the angle B'AD' with the line B'D', and we conclude that $\frac{D^{\prime}N^{\prime}}{N^{\prime}B^{\prime}}=\frac{AD^{\prime}}{AB^{\prime}}$. Finally, since the point K' is the harmonic conjugate of the point A with respect to the segment C'D', we have $\frac{C^{\prime}K^{\prime}}{K^{\prime}D^{\prime}}=-\frac{C^{\prime}A}{AD^{\prime}}$. Thus, $\frac{D^{\prime}N^{\prime}}{N^{\prime}B^{\prime}}\cdot\frac{B^{\prime}M^{\prime}}{M^{\prime}C^{\prime}}\cdot\frac{C^{\prime}K^{\prime}}{K^{\prime}D^{\prime}}=\frac{AD^{\prime}}{AB^{\prime}}\cdot\frac{AB^{\prime}}{C^{\prime}A}\cdot\left(-\frac{C^{\prime}A}{AD^{\prime}}\right)=-1$. Hence, by the Menelaos theorem, the points M', N' and K' lie on one line. The problem is solved. EDIT: This problem comes from the 24th Russian Mathematical Olympiad 1997. Darij

As I remember a more general problem is G8 in the shortlist of IMO2002. I can post the original solution if it is needed. [Moderator edit: Thanks for noticing! Now I have found various solutions for this problem G8 in the following sources: http://www.kalva.demon.co.uk/short/soln/sh02g8.html http://www.mathlinks.ro/Forum/viewtopic.php?t=22165 http://www.mathlinks.ro/Forum/viewtopic.php?t=22201 http://www.mathlinks.ro/Forum/viewtopic.php?t=17322 http://www.mathlinks.ro/Forum/viewtopic.php?t=15588 post #2 file IMO_2002_shortlist.pdf page 21 .]

Let $M'$ and $N'$ be the midpoints of $BC$ and $BD$ respectively. It is obvious that $\triangle MM'K\sim\triangle KN'N$. Then $\angle MKC+\angle NKD=\angle BDC+\angle BCD+(\angle MKM'+\angle NKN')=\angle BDC+\angle BCD+\pi-\angle KM'M=\frac{\pi}{2}$. The result follows.

Please clarify your solution . I think It's not obvious and I believe there is a simpler solution

Basically my solution is the same as the one posted by E. Leung (even the notation!), so I'm not going to explain it in details. Anyway, although it's not really that obvious, I personally believe that anyone associated with Math Olympiad should be able to solve it.

A equivalent enunciation of the proposed problem: Let the triangle $ABC$, the midpoint $D$ of the side $[BC]$ and two isosceles triangles $ABM,\ ACN\ (MA=MB,\ NA=NC)$ such that the line $AB$ separes the points $D,M$, the line $AC$ separates the points $D,N$ and $m(\widehat {AMB})+m(\widehat {ANC})=180^{\circ}.$ Prove that $DM\perp DN$ and $m(\widehat {DMN})=\frac 12 m(\widehat {AMB})$. Remark. After at least a hour I will prove with complex numbers !

My solution was somewhat different, but I also used spiral similarity. [I am just solving the problem the way levi stated it.] Let $C'$ and $D'$ be the reflections of $C$ and $D$ in $M$ and $N$ respectively. It is easy to see that the right-angled triangles $BCC'$ and $BD'D$ are similar (this follows from $\angle{BMC} + \angle{BND} = 180^{\circ}$). Hence, triangles $BC'D$ and $BCD'$ are similar, and there is a spiral similarity (angle $90^{\circ}$) mapping the $BC'D$ to $BCD'$. This spiral similarity maps $C'D$ to $CD'$, so $C'D$ and $CD'$ are perpendicular. Since $MK \parallel C'D$, $NK \parallel CD'$ we are done.

(See the last my reply). I note: $ u=m(\widehat {AMB}),\ v=m(\widehat {ANC})\ (u+v=180^{\circ}),\ x=\cos u+i\cdot \sin u,\ y $ $ =\cos v+i\cdot \sin v\ (xy=-1). $ $ A(a),\ B(b),\ C(c),\ D(d),\ M(m),\ N(n)\Longrightarrow 2d=b+c,\ m= $ $ \frac {a-bx}{1-x},\ n=\frac {c-ay}{1-y}\Longrightarrow $ $ m-d=\frac {2a-b(1+x)-c(1-x)}{2(1-x)},\ n-d=\frac {-2ay-b(1-y)+c(1+y)}{2(1-y)} $ $ ,\ xy=-1\Longrightarrow $ $ n-d=\frac {2a\cdot \frac 1x -b(1+\frac 1x )+c(1-\frac 1x)}{2(1+\frac 1x)}=\frac {2a-b(1+x)-c(1-x)}{2(1+x)}\Longrightarrow $ $ \frac {n-d}{m-d}=\frac {1-x}{1+x}=\frac {1-\cos u-i\cdot \sin u}{1+\cos u +i\cdot \sin u}=\frac {2\sin^2 \frac u2 -2i\sin \frac u2\cos \frac u2}{2\cos^2 \frac u2 +2i\sin \frac u2 \cos \frac u2}=-i\cdot \tan \frac u2 \Longrightarrow $ $ (m-d)=i\cdot cot \frac u2 \cdot (n-d)\Longrightarrow DM\perp DN, \tan \frac u2=\frac {DN}{DM} $, i.e. $ m(\widehat {DMN})=\frac 12 \cdot m(\widehat {AMB}). $ 1. Remark. Let $ P $ be the point of the line $ BC $ for which $ m(\widehat {APC})=u,\ m(\widehat {APB})=v\ (u+v=360^{\circ}) $. Then the $ AMBM,\ ANCP $ are cyclic quadrilaterals and thus, we obtain the initial proposed problem !. 2. Remark. This problem admits a generalization (see http://www.mathlinks.ro/Forum/viewtopic.php?p=289313#p289313 ): $ \blacksquare $. Let $ ABC $ be a triangle and three isosceles triangles $ BCL,\ CAN,\ ABM\ (LB=LC,NC=NA,MA=MB) $ such that the lines $ BC,CA,AB $ separe the pointa $ A,B,C $ from the points $ L,N,M $ respectively and $ m(\widehat {BLC})+m(\widehat {CNA})+m(\widehat {AMB})=360^{\circ} $. Then $ m(\widehat {MLN})=\frac 12\cdot m(\widehat {BLC}),\ m(\widehat {LNM})=\frac 12 \cdot m(\widehat {CNA}),\ m(\widehat {NML})=\frac 12\cdot m(\widehat {AMB}).\blacksquare $. For $ L: =D $ we obtain the previous problem !.

Dear Mathlinkers, 1. M' the symmetric of M wrt K 2. the triangles BMN and DM'N are equal and we are done Sincerely Jean-Louis

Hi J.L. jayme wrote: 2. the triangles BMN and DM'N are equal I only can see that BN=ND and MN=M'N, but somewhat I can't go any further. jayme wrote: and we are done why? If you would be so kindly to explain a little bit longer... I would be even more grateful.

Dear Mathlinkers, <MBN = <MBA + <ABN ; <MBN = <ACM + <NDA ; <ACM = <CDM' = <ADM' ; <MBN = <ADM' + <NDA ; <MBJN = <NDM'. Sincerely Jean-Louis

I have a different one using transformations. Let $\angle CAB = A$ and $\angle DAB = 180 - A$. Also, note that $CB/CM = 2 \cos A/2$(by using sine law and $\sin(A) = 2\sin A/2 \cos A/2$. Consider the sequence of transformations: (1) A spiral similarity with centre $C$, ratio $2\cos A/2$ and angle $A/2$. (2) A spiral similarity with centre $D$, ratio $\dfrac{1}{2 \sin A/2}$ and angle $90-A/2$. Note that this takes $M \to B, B \to N$, so $M \to N$. Apply the first transformation to $K$, let the image be $K'$. Now, contruct a right angled triangle $\triangle ECD$ with hypotenuse $CD$, such that $\angle ECD = A/2$. Now, By the dilation properties of (1) $CK' = BC \cos A/2$. But by our right triangle, $CE = BC \cos A/2$, so $E = K'$. Then the second transformation takes $E$ to $E'$, where $E'$ lies on $CD$ (by our right triangle). Also by our right triangle, $DE = CD \sin A/2 = 2DK \sin A/2$, so in fact $K = E'$, meaning $K$ stays invariant under the transformations. So, the resultant transformation is a spiral similarity about $K$ with net rotation $(90 - A/2) + A/2 = 90$ and dilation $\dfrac{1}{\tan A/2}$, which takes $M \to N$ i.e. $\angle MKN = 90$. Remark: There is a shorter solution given by Number1. Consider the composition $R_M \circ R_N$ taking $C \to B \to D$. This is equivalent to a half turn about $K$. But, then we know that $\angle (KM, MN) = \angle CMB/2$ and $\angle (KN, MN) = \angle DMB/2$. Easy to see $\angle CMB + \angle DMB = 180$, so $\angle MKN = 90$.