Transformation a quadrilateral to a rectangle.

Letbe a quadrilateral. .is an arbitrary line parallel to..be an arbitrary point on the circle with diameter.cuts the line throughand parallel toatO.. Prove that the line throughand perpendicular tocuts the line throughand perpendicular toat a point on.Solution 1.- Consider the homology fixing the penciland the lineIt carries the figureinto the figureThe infinite point of the linegoes then to the intersectionofwith the parallel throughtoparallelfromtois therefore the limiting line of the figureAs a result,goes to the pointwhere the parallels fromandtoandintersect, in other words the intersection of the perpendicular toatand the perpendicular toatlies onComment: Can you explain to me some concepts: homology, infinite point, limiting line? Thank you in advanceAnswer: The concept of point at infinity is exactly the same thing as direction in the plane, e.g. a set of parallel lines (having the same direction) is a set of lines passing through a point at infinity. All points at infinity lie on the line at infinity. The Euclidean plane with this line added is known as projective plane. So here, any two lines always intersect, parallels are no exception.A bijection between two figuressuch that a point and an indicent line go to a point a an indicent line, is called homography. The limiting line is simply the image of the line at infinity under a homographic transformation. Hence, in general there are 2 limiting linesin a bijection, one for each figureA homology is a homography that fixes a line.

PQ is parallel to BC-2013 China TST Quiz 1 Day 1 P1

The quadrilateralis inscribed in circle.is the intersection point ofand.andmeet at. Let the projection ofonandbeand, respectively. Letandbe the midpoints ofand, respectively. If the circumcircle ofonly meets segmentat, and the circumcircle ofonly meets segmentat, prove thatis parallel to.Solution 1.- Let the midpoints of sidesandbeandI will prove thatand. If this is proved we are done.First we observe that lie on a circle centered at) andfor the same reason..Now letbe the reflection ofoveris a parallelogram. Observe that and Now we make the important observation that the two figuresandare homothetic centeringmappingtototoand Nto. Hence proving thatare concyclic .Done!Comment 1: what do they mean by meet only at p and q? like its tangent? or meet the segmentAnswer to comment: meet at segmentSolution 2: Actually, it's very simple if you can see that=,=(Stewart's theorem),=, triangleis congruent to triangleand. Then||and Q is the midpoint of. Similarly, on the other side||and P is the midpoint of. Thus||.Solution 2.- Letbe the midpoints of, andthe circumcircle of,the intersection ofwith. Then , solies onas well., so, which means, so.

Solution 3.- Letbe the midpoints of, andthe circumcircle of,the intersection ofwith. Then , solies onas well., so, which means, so.Euler lineLet ABC be a scalene triangle andthe reflections ofonto..Let D be the intersection of.Prove that thatis parallel to Euler lineSolution.- First we showOis onAD.concyclic.Now letmeetatconcyclic, and hence.

Antigonal conjugate of I wrt its cevian triangle lies on O'ILetbe a triangle and letbe its incenter. Let,,be the intersections of,,with the sides,, and. Prove that the antigonal conjugate ofwith respect to trianglelies on the line, whereis the circumcenter of triangle.Solution 1.- Hope this will be helpfulRewriteas.Letbe the projection ofon.Letbe the projection ofon.Letbe the midpoint of.Invert with respect toand denoteas the image ofunder this inversion.Easy to seeandis the midpoint of.Sinceall lie on,so are collinear andis the projection ofon. Similarly, we can getis the projection ofon. Since, sois a harmonic quadrilateral. ie.is the midpoint of.Similarly, we can getis the midpoint of.From the definition ofwe can getis the reflection ofwith respect to.Sincepass through the midpoint of,sois the nine point circle of triangle. ie.is the nine point center of triangle. Since the Antigonal conjugate ofis the reflection ofin the Poncelet pointofandlie on ,so we get the image of the Antigonal conjugate ofis the midpoint ofandis the common point ofwhich is the Anti-steiner point of Euler line (called Kiepert focus which isin ETC ) of triangle. (I use a property of Anti-steiner point here: Ifis the reflection ofwith respect to, then concur at the Anti-steiner point of the line pass throughand the orthocenter of trianglewith respect to triangle). Sinceis the pedal circle of, so the image oflie on the line pass throughand the Kosnita point of triangle( Kosnita point is the isogonal conjugate of the nine point center which isin ETC ). Now we only have the showare collinear ... this is a well known fact but I don't have the synthetic proof yetComment: Now we only have to prove the following theorem :Theorem:Letbe the nine point center of.Letbe the Kosnita point of.Letbe the Kiepert focus of.Thenare collinearProof:Letbe the circumcenter of.Letbe the orthocenter of.Letbe the tangential triangle of.Letbe the midpoint of, respectively .Letbe the reflection ofin, respectively .Sinceis the Anti-steiner point of the Euler line of, sois the Feuerbach point of. Sinceis the circumcenter of, respectively . So from the property of the Kosnita point we get. Consider the homothetic with centerand factor, then ,hence the imageofunder this homothetic isof.Sinceis the incenter of, so the imageofisof(Antigonal conjugate of the incenter). Sinceis theof,so fromhttp://www.artofproblemsolving.com/Foru ... 6&t=612271we getare collinear, henceare collinear .Q.E.DProblem of conic and three circles!Letare concyclic;are concyclic;are concyclic. Letare on conic. Prove that:are concyclic.Answer.- Labelthe given conic with axesArbitrary circlew throughA, DcutsCagain atare points onsuch that the tangents ofatare parallel torespectively. Letbe the intersection of these tangents andBy generalized power of point, we have

Hence,is either onorandare equal inclined toThus whenvaries, keepingfixed, all linesgo through a fixed direction. As a result, we deduce thatandIfcutsagain atthenare concyclic.An inequality in a squareI research a pure geometric solution without any calculation1. ABCD a square2. E a point on the segment AD3. F the foot of the perpendicular to CE through B4. M the midpoint of EF5. G the point of intersection of the parallel to BC through M with the perpendicular bissector of BF.Prove : AC < 2.FG.Sincerely, Jean-LouisAnswer.- Courtesy of Clarence Chew:We want to show that: 2BC2=AC2