Phew. I've been working on this one for while and still can't get anywhere.

Vectors! Take the origin to be the circumcenter of $ ABCD$. $ X=(1-\alpha) A + \alpha B$ and \[ |C-D|=|X-A|=\alpha|B-A|\] So, \[ X = A + \frac{(B-A)}{|B-A|}|C-D|\] Likewise, \[ Y = C + \frac{(B-C)}{|B-C|}|A-D|\] First: as $ A,B,C,D$ are concyclic, $ \angle(AD,CD)=180-\angle(AB,CB)$ Thus, \[ \frac{(B-A)\cdot(B-C)}{|B-A||B-C|}=-\frac{(D-A)\cdot(D-A)}{|D-A||D-C|}\qquad (1)\] We wish to prove that $ A,\frac{X+Y}{2},\text{ and }C$ form a right triangle - that is, \[ \left(\frac{X+Y}{2}-A\right)\cdot\left(\frac{X+Y}{2}-C\right)=0\] This is clearly equivalent to:\[ \left(\frac{(B-A)}{|B-A|}|C-D|+\frac{(B-C)}{|B-C|}|A-D|\right)^2=(A-C)^2\] Let's expand the left side and appeal to $ (1)$. \[ \begin{eqnarray*}\left(\frac{(B-A)}{|B-A|}|C-D|+\frac{(B-C)}{|B-C|}|A-D|\right)^2 &=& |C-D|^2+|A-D|^2+ \ 2\left[\frac{(B-A)\cdot(B-C)}{|B-A||B-C|}\right]|C-D||A-D|\\ &=& |C-D|^2+|A-D|^2 - 2(C-D)\cdot(A-D)\qquad\text{from }(1) \\ &=& \left( (C-D) - (A-D) \right)^2 = (A-C)^2\] which therefore establishes that $ AM$ is perpendicular to $ AC$, i.e. the goal.

Let $ A'$ be the reflection of $ A$ with respect to $ M$. In this case, the quadrilateral $ AXA'Y$ is a parallelogram and therefore $ YA'=AX=CD$, $ CY = AD$ and $ \angle{A'YC}=180^{\circ} - \angle{A'YB} = 180^{\circ} -\angle{CBA} = \angle{CDA}$. Thus, the triangles $ ACD$ and $ A'CY$ are congruent, hence $ AC = A'C$, and since $ M$ is the midpoint of $ AA'$, it follows that $ CM \perp AM$.

http://www.mathlinks.ro/viewtopic.php?p=1156448 http://www.mathlinks.ro/viewtopic.php?t=85214

I'm posting my solution, which I've sent to bboypa yesterday, just to show how two solutions of two different people can be similar, even in wording: Let $ A'$ be the reflection of the point $ A$ through the point $ M$. Then quadrilateral $ AXA'Y$ is a parallelogram since its diagonals bisects each other. Therefore $ YA' = AX = CD$ and $ YA' \parallel AX$. It follows that $ \angle CYA' = 180^{\circ} - \angle ABC = \angle ADC$ since quadrilateral $ ABCD$ is cyclic. The given condition $ CY = AD$ implies that triangles $ CYA'$ and $ ADC$ are congruent (Side - Angle - Side). It means that $ CA' = CA$ and the triangle $ ACA'$ is isosceles. Since $ AM=MA'$ it follows that $ CM \perp AM$ and the proof is finished.

TomciO wrote: I'm posting my solution, which I've sent to bboypa yesterday, just to show how two solutions of two different people can be similar, even in wording: Let $ A'$ be the reflection of the point $ A$ through the point $ M$. Then quadrilateral $ AXA'Y$ is a parallelogram since its diagonals bisects each other. Therefore $ YA' = AX = CD$ and $ YA' \parallel AX$. It follows that $ \angle CYA' = 180^{\circ} - \angle ABC = \angle ADC$ since quadrilateral $ ABCD$ is cyclic. The given condition $ CY = AD$ implies that triangles $ CYA'$ and $ ADC$ are congruent (Side - Angle - Side). It means that $ CA' = CA$ and the triangle $ ACA'$ is isosceles. Since $ AM = MA'$ it follows that $ CM \perp AM$ and the proof is finished. Excellent solution Tomci0 & pohoatza! @pohoatza..i'll attend also you next round if you have enough time

I proved that pythagorean theorem is true in the triangle $ \triangle AMC$ by using AL-KACHI and median theorems. I don't know reflections, can someone give me a link? -- Wikipedia isn't very clear, i found.

I've just noticed that you are talking about what i call "central symetrie". ^^"

Seeing Pohoatza's, one could think: "What else could be added?" However, here's my approach: Name K and L the intersection of AB and CD, respectively BC and AD. Also name N and P the middle of AD and CX respectively. I shall use the following properties: 1) A quadrilateral having 2 opposite sides equal, has the bisector of the angle made by them parallel to the line joining the midpoints of the other 2 sides 2) In a cyclic quadrilateral KABCLD, the bisectors of the angles BKC and BLA are perpendicular. (I do not see necessary to prove the above statements, they are well known) According to (1), NP is parallel to the inner angle bisector of <BKC, so we intend to prove AM||NP, or AMPN an isosceles trapezoid. It has AN=MP, both being half of AD, and MP||CY. If the angle bisector of <BKC cuts AD and BC at N' and P', then the angles AKN' and BKP' are supplementary (add 180 deg), the triangles KAD and KCB being similar, therefore <ANP = <NPM and AMPN is isosceles trapezoid, consequently AM ||NP||KP'. Same way CM is parallel to the other angle bisector. Best regards, sunken rock