Because of the cyclic quadrilateral $ ABCD,$ it is easy to see the configuration of the triangle $ \bigtriangleup OEF$ and the similar right triangles $ \bigtriangleup EBO\ \bigtriangleup FCO,$ erected ( outwardly to it ) on its side-segments $ OE,\ OF,$ respectively. It is well known that the midpoint $ M$ of the side-segment $ BC,$ lies on the midperpendicular of the segment $ EF.$ Similarly the midpoint $ N$ of the side-segment $ AD,$ lies also on the same line ( we consider now, the similar right triangles $ \bigtriangleup EAO\ \bigtriangleup FDO,$ erected inwardly to it, on the same side-segments ). That is we have that the line segment $ MN,$ is the midperpendicular of the segment $ EF$ and so, from the congruent triangles $ \bigtriangleup EMN,\ \bigtriangleup FMN,$ we have $ (EM)\cdot (FN) = (EN)\cdot (FM)$ $ ,(1)$ and the proof of the part (1), is completed. $ \bullet$ As for the part (2) of the proposed problem, we can say that it is not true in general. If we consider two points $ B',\ C',$ on $ AB,\ CD$ respectively, such that $ \angle EOB' = \angle FOC'$ then, the midpoint $ M'$ of the segment $ B'C',$ lies also on the line segment $ MN$ as the midperpendicular of $ EF$ and although the relation $ (EM')\cdot (FN) = (EN)\cdot (FM')$ is true, however the quadrilateral $ AB'C'D,$ is not cyclic $ ($ because $ B'C'$ is not parallel to $ BC$ $ ).$ But clearly, when the given triangle $ \bigtriangleup ABC$ is isosceles, the quadrilateral $ AB'C'D$ as before is cyclic and so, in this particular case, the part (2) of the proposed problem is true. Kostas Vittas.

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t=250255.pdf (6kb)

as for (1) Denote R,S,T,Q are midpoints of GD,GA,GB,GC , respectively △NRF≌△ESN △ETM≌△MQF So EN=FN EM=FM => EM*FN=EN*FM

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Can someone explain to me the solution of part b. In vittasko's solution O isn't the intersection of AC' and B'D, nor is it the intersection of AB' and DC'.

The first part of this problem is in fact a past olympiad problem. I saw it few years ago, but we are required to prove $ EM=FM$ instead of $ (EM)\cdot (FN) = (EN)\cdot (FM)$. For the second part, let $ K$ be the intersection of $ NS$ and $ MT$ (using the configuration in plane geometry' post), then $ KEST$ is a cyclic quadrilateral. Thus $ \angle NSE=\angle ETM$ no matter $ ABCD$ is cyclic or not. Then by cosine law, it can be shown that $ (OA)\cdot (OB) = (OC)\cdot (OD)$ also implys $ (EM)\cdot (FN) = (EN)\cdot (FM)$. And it is easy for us to construct a triangle with this property

thecool wrote: Can someone explain to me the solution of part b. In vittasko's solution O isn't the intersection of AC' and B'D, nor is it the intersection of AB' and DC'. I'm confused with vittasko's solution as well. The point $ O$ changes when you change $ B$ to $ B'$ and $ C$ to $ C'$...

I am sorry dear thecool and jbmorgan and all my friends, because of the confusion from all what I wrote about the solution of the part (2) of the proposed problem. Although we can say that it is not true in general, however I used a false argument ( as for the point $ O\equiv AC\cap BD$ ) in trying to prove it. It is better to use the argument of an arbitrary (non isosceles) trapezium $ ABCD$ with $ AD\parallel BC,$ where is true the relation as the problem states (easy to prove) and clearly, it is not a cyclic quadrilateral. I will post here later more details. Give please a little time. Sorry again, Kostas Vittas.

vittasko wrote: …It is well known that the midpoint $ M$ of the side-segment $ BC,$ lies on the midperpendicular of the segment $ EF.$ LEMMA 1. - A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABD,\ \bigtriangleup ACE$ be, two similar right triangles with $ \angle ABD = \angle ACE = 90^{o}$ and $ \angle BDA = \angle CEA = \angle \omega,$ erected outwardly (or inwardly) to it. Prove that the midperpendicular of the side-segment $ BC,$ also bisectw the segment $ DE.$ PROOF. - (we used here, the same argument as mentioned before, by plane geometry ). Let $ N$ be, the midpoint of the segment $ DE$ and it is enough to prove that $ NB = NC$ $ ,(1)$ We denote as $ K,\ L,\ M,$ the midpoints of the segments $ AD,\ AE,\ BC$ respectively and is is easy to show that $ KN = \frac {AE}{2} = LC$ $ ,(2)$ and $ LN = \frac {AD}{2} = KB$ $ ,(3)$ from $ KN\parallel AE,\ LN\parallel AD$ and the right triangles $ \bigtriangleup ADB,\ \bigtriangleup ACE.$ From$ (2),$ $ (3)$ and because of $ \angle BKN = \angle BKA + \angle AKN = \angle CLA + \angle ALN = \angle CLN,$ we conclude that the triangles $ \bigtriangleup KBN,\ \bigtriangleup LNC$ are congruent. Hence, we have that $ NB = NC$ $ ,(4)$ and the proof of the Lemma is completed. NOTE : As an interesting remark in the configuration of the above Lemma 1, we can say that $ \angle BNC = \angle 2\omega.$ Kostas Vittas.

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t=250255(a).pdf (5kb)

vittasko wrote: …It is better to use the argument of an arbitrary (non isosceles) trapezium $ ABCD$ with $ AD\parallel BC,$ where is true the relation as the problem states (easy to prove) and clearly, it is not a cyclic quadrilateral. LEMMA 2. - An arbitrary trapezium $ ABCD$ is given with $ AD\parallel BC$ and let be the point $ O\equiv AC\cap BD.$ We denote as $ M,\ N,$ the midpoints of its side-segments $ BC,\ AD$ respectively and let $ E,\ F$ be, the orthogonal projections of $ O,$ on $ AB,\ AC,$ respectively. Prove that $ (EN)\cdot (FM) = (EM)\cdot (FN).$ PROOF. - Let be the point $ P\equiv AB\cap CD$ and it is easy to show that the points $ M,\ N,\ P$ are collinear, applying the Thales theorem. From the complete quadrilateral $ PAODBC,$ we have that the points $ P,\ N,\ O,\ M,$ are in harmonic conjugation as well. So, we conclude that the pencil $ E.PNOM,$ is also in harmonic conjucation. Because of now, $ EP\perp EO,$ we conclude that the line segment $ EO$ is the angle bisector of the angle $ \angle MEN$ and then, we have $ \frac {EM}{EN} = \frac {OM}{ON}$ $ ,(1)$ Similarly, the line segment $ FO$ is the angle bisector of the angle $ \angle MFN$ and then, we have also $ \frac {FM}{FN} = \frac {OM}{ON}$ $ ,(2)$ From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {EM}{EN} = \frac {FM}{FN}$ $ \Longrightarrow$ $ (EN)\cdot (FM) = (EM)\cdot (FN)$ and the proof is completed. Kostas Vittas.

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t=250255(b).pdf (3kb)

vittasko wrote: …NOTE : As an interesting remark in the configuration of the above Lemma 1, we can say that $ \angle BNC = \angle 2\omega.$ PROOF. - We draw the circles $ (K),\ (L),$ with diameters $ AD,\ AE$ respectively and let be the points $ B'\equiv (K)\cap BC$ and $ C'\equiv (L)\cap BC.$ We denote as $ D',\ K',\ A',\ L',\ E',$ the orthogonal projections of the points $ D,\ K,\ A,\ L,\ E$ respectively and it is easy to show that $ D'B' = BA'$ $ ,(1)$ because of the point $ K'$ is the midpoint of the segments $ BB'$ and $ A'D'.$ Similarly, we have that $ A'C = C'E'$ $ ,(2)$ because of the point $ L'$ is the midpoint of the segments $ A'E'$ and $ CC'.$ The right triangles $ \bigtriangleup D'B'D,\ \bigtriangleup E'C'E$ are similar, because they have $ \angle D'B'D = \angle BAD = 90^{o} - \angle \omega = \angle CAE = \angle E'C'E$ and so, we conclude that $ \frac {D'B'}{E'C'} = \frac {D'D}{E'E}$ $ ,(3)$ From $ (3)$ $ \Longrightarrow$ $ \frac {D'B}{D'B + E'C'} = \frac {D'D}{D'D + E'E}$ $ \Longrightarrow$ $ \frac {D'B'}{2(MB)} = \frac {D'D}{2(MN)}$ $ ,(4)$ $ ($ from $ D'B' + C'E' = BA' + A'C = BC = 2(MB)$ and $ D'D + E'E = 2(MN)$ $ ).$ From $ (4)$ $ \Longrightarrow$ $ \frac {D'B'}{MB} = \frac {D'D}{MN}$ and so, we conclude that the right triangles $ \bigtriangleup D'B'D,\ \bigtriangleup MBN,$ are similar. Hence, we have that $ \angle MBN = \angle D'B'D = 90^{o} - \angle \omega$ $ ,(5)$ From $ (5)$ $ \Longrightarrow$ $ \angle BNC = \angle 2\omega$ and the proof is completed. Kostas Vittas.

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t=250255(c).pdf (10kb)

Here is my solution Call P is midpoint of PO. By Gauss line theory, we have M, N, J is conlinear. Similar TST USA 2000, we get EF is perpendicular to MN, in the other hand, JE = JF. Hence, JMN is perpendicular bisector of MN……..