bomb wrote: The diagonals AC and BD of a cyclic ABCD interesct at E. Let O be circumcentre of ABCD. If midpoints of AB, CD, OE are collinear prove that AD=BC. This seemingly obvious proposition is incorrect, which is why it is difficult to prove. If $AD = BC$, i.e., the cyclic quadrilateral is an isosceles trapezoid, the points $O, E$ and the midpoints of the segments $AB, BC, OE$ are all collinear. But the midpoints of $AB, CD, OE$ are also collinear, if the cyclic quadrilateral has perpendicular diagonals, and such cyclic quadrilateral does not have to be an isosceles trapezoid. By 2nd Brahmagupta's theorem, the diagonal intersection of a cyclic quadrilateral with perpendicular diagonals is identical with its anticenter (the intersection of its maltitudes). The midpoint $G$ of the circumcenter-anticenter segment is the centroid of any cyclic quadrilateral. The centroid is the intersection of the diagonals $KM, LN$ of the parallelogram $KLMN$, where $K, L, M, N$ are the midpoints of the sides $AB, BC, CD, DA$.

yetti wrote: bomb wrote: The diagonals AC and BD of a cyclic ABCD interesct at E. Let O be circumcentre of ABCD. If midpoints of AB, CD, OE are collinear prove that AD=BC. This seemingly obvious proposition is incorrect, which is why it is difficult to prove. If $AD = BC$, i.e., the cyclic quadrilateral is an isosceles trapezoid, the points $O, E$ and the midpoints of the segments $AB, BC, OE$ are all collinear. But the midpoints of $AB, CD, OE$ are also collinear, if the cyclic quadrilateral has perpendicular diagonals, What is more interesting (because much harder to prove) is that the converse also holds, i. e. we can prove: Theorem 1. Let ABCD be a (not necessarily convex) cyclic quadrilateral with circumcenter O, and let the diagonals AC and BD of the quadrilateral ABCD intersect at a point E. Then, the midpoints of the segments AB, CD and OE are collinear if and only if either AD = BC or $AC\perp BD$. Now, here is a proof of Theorem 1. We will use directed angles modulo 180º. First, note that two chords of a circle are equal if and only if their chordal angles are equal or oppositely equal; applying this to the chords AD and BC of the circumcircle of the quadrilateral ABCD, we see that AD = BC holds if and only if either < ABD = < BDC or < ABD = - < BDC. Now, since the quadrilateral ABCD is cyclic, < BDC = < BAC, and thus the equation < ABD = < BDC becomes < ABD = < BAC, what is equivalent to AC || BD. On the other hand, the equation < ABD = - < BDC rewrites as < ABD = < CDB, what is equivalent to AB || CD. Hence, we have AD = BC if and only if AC || BD or AB || CD. - If AC || BD, then the quadrilateral ABDC is a (self-intersecting) trapezoid; hence, the point E, being the point of intersection of the parallel lines AC and BD, must be an infinite point, so the midpoint of the segment OE is the point E, and thus we easily see that the midpoints of the segments AB, CD and OE are collinear (in fact, this is equivalent to stating that the line joining the midpoints of the segments AB and CD passes through the infinite point E, i. e. is parallel to the lines AC and BD; but this follows from the fact that the quadrilateral ABDC is a trapezoid). - If AB || CD, then the quadrilateral ABCD is an isosceles trapezoid (trapezoid since AB || CD, isosceles since a cyclic trapezoid is always isosceles), hence it is symmetric with respect to an axis, and thus the midpoints of the segments AB, CD and OE are collinear (they all lie on this axis). Thus, we see that if AD = BC, then the midpoints of the segments AB, CD and OE are collinear. So Theorem 1 is proven for the case when AD = BC. It remains to consider the case when $AD\neq BC$. So, well, assume that $AD\neq BC$. Since we have shown in the above that we have AD = BC if and only if AC || BD or AB || CD, we now see that, as we assumed that $AD\neq BC$, we have $AC\not\parallel BD$ and $AB\not\parallel CD$. Since $AC\not\parallel BD$, the point of intersection E of the lines AC and BD is a finite point. Since $AB\not\parallel CD$, the point of intersection L of the lines AB and CD is a finite point. Let M and N be the midpoints of the segments AB and CD. Then, we have to show that the points M and N and the midpoint of the segment OE are collinear if and only if either AD = BC or $AC\perp BD$. Well, since we have assumed that $AD\neq BC$, this simplifies to proving that the points M and N and the midpoint of the segment OE are collinear if and only if $AC\perp BD$. So we have to establish two assertions: Assertion 1. If $AC\perp BD$, then the points M and N and the midpoint of the segment OE are collinear. Assertion 2. If the points M and N and the midpoint of the segment OE are collinear, then $AC\perp BD$. A proof of Assertion 1 was given by Yetti in post #2. Another proof goes as follows: Since $AC\perp BD$, we have < AEB = 90º, so that the triangle AEB is right-angled at E. The point M, being the midpoint of its hypotenuse AB, must be its circumcenter. Hence, MB = ME, so that the triangle BME is isosceles, and thus < MEB = < EBM. In other words, < (ME; BD) = < DBA. Since the quadrilateral ABCD is cyclic, < DBA = < DCA, so that < (ME; BD) = < DCA = < (CD; AC). On the other hand, the point O, being the circumcenter of the cyclic quadrilateral ABCD, must lie on the perpendicular bisector of its side CD; hence, $ON\perp CD$, since N is the midpoint of the segment CD. On the other hand, $AC\perp BD$ yields < (BD; AC) = 90º. Hence, < (ME; CD) = < (ME; BD) + < (BD; CD) = < (CD; AC) + < (BD; CD) = < (BD; AC) = 90º, so that $ME\perp CD$, what, compared with $ON\perp CD$, yields ME || ON. Similarly, NE || OM. Hence, the quadrilateral MENO is a parallelogram, and thus its diagonals MN and OE bisect each other. Particularly, this means that the midpoint of the diagonal OE lies on the diagonal MN, i. e. the points M and N and the midpoint of the segment OE are collinear. Hence, Assertion 1 is proven. Now we come to the hard part, namely the proof of Assertion 2: Assume that the points M and N and the midpoint of the segment OE are collinear. In other words, the midpoint T of the segment OE lies on the line MN. Let the lines BC and DA intersect at a point F (this point F can possibly be infinite), and let R be the midpoint of the segment EF. The segments AB, CD and EF are the diagonals of the complete quadrilateral formed by the lines BC, DA, AC and BD. Hence, the midpoints M, N and R of these diagonals are collinear. In other words, the point R lies on the line MN. Now, since the points R and T are the midpoints of the sides EF and OE of the triangle OEF, we have RT || OF. Since the points R and T both lie on the line MN, this becomes MN || OF. Now, it is a well-known fact that if ABCD is a cyclic quadrilateral, then the line joining the points $AB\cap CD$ and $BC\cap DA$ is the polar of the point $AC\cap BD$ with respect to the circumcircle of the cyclic quadrilateral ABCD (this was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=38871 (problem 2), http://www.mathlinks.ro/Forum/viewtopic.php?t=25036 , http://www.mathlinks.ro/Forum/viewtopic.php?t=5326 , http://www.mathlinks.ro/Forum/viewtopic.php?t=550 - I have just copied the links from my latest post). Applying this fact to the cyclic quadrilateral CABD instead of ABCD, we obtain that the line joining the points $CA\cap BD$ and $AB\cap DC$ is the polar of the point $CB\cap AD$ with respect to the circumcircle of the cyclic quadrilateral CABD (which is obviously the same as the circumcircle of the cyclic quadrilateral ABCD). In other words, the line EL is the polar of the point F with respect to the circumcircle of the cyclic quadrilateral ABCD. Since the center of this circumcircle is O, we thus have $EL\perp OF$. Hence, MN || OF becomes $MN\perp EL$, so that < (EL; MN) = 90º. Now, let U and V be the orthogonal projections of the point E on the lines AB and CD. Then, < LUE = 90º and < LVE = 90º, so the points U and V lie on the circle with diameter LE. Hence, < LUV = < LEV. On the other hand, $EV\perp CD$ yields < (EV; CD) = 90º, and thus < LUV = < LEV = < (EL; EV) = < (EL; CD) - < (EV; CD) = < (EL; CD) - 90º = < (EL; CD) - < (EL; MN) = < (MN; CD). In other words, < MUV = < MNV. Hence, the points U, V, M and N lie on one circle. The center of this circle lies on the perpendicular bisectors of the segments MN, MU and NV (as the center of a circle lies on the perpendicular bisector of any chord of the circle). Hence, we see that the common point of the perpendicular bisectors of the segments MU and NV (these two perpendicular bisectors have only one common point, since they are not parallel, since $MU\not\parallel NV$, what is equivalent to $AB\not\parallel CD$) lies on the perpendicular bisector of the segment MN. Note that $ON\perp CD$ (as we showed in the proof of Assertion 1). On the other hand, $EV\perp CD$, and thus ON || EV. Hence, the quadrilateral OEVN is a trapezoid. Hence, the line joining the midpoints of its legs OE and NV must be parallel to its bases ON and EV, and thus perpendicular to the line CD (since these bases ON and EV are perpendicular to CD). In other words, the midpoint T of the leg OE must lie on the perpendicular to the line CD at the midpoint of the leg NV. But, obviously, the perpendicular to the line CD at the midpoint of the leg NV is simply the perpendicular bisector of the segment NV. Hence, the point T lies on the perpendicular bisector of the segment NV. Similarly, the point T lies on the perpendicular bisector of the segment MU. Thus, the point T is the common point of the perpendicular bisectors of the segments MU and NV. Thus, as we showed above, the point T must lie on the perpendicular bisector of the segment MN. But we know that the point T lies on the line MN. Hence, the point T must be the midpoint of the segment MN. So the point T is the midpoint of both segments MN and OE. Hence, the quadrilateral MENO is a parallelogram (as its diagonals MN and OE bisect each other). If AE = BE and CE = DE, then $\dfrac{AE}{CE}=\dfrac{BE}{DE}$, and thus, by Thales, AB || CD, what contradicts with the supposition $AB\not\parallel CD$. Hence, at least one of the equations AE = BE and CE = DE must be wrong. WLOG, assume that the equation AE = BE is wrong, i. e. that $AE\neq BE$. Since the quadrilateral MENO is a parallelogram, we have ME || ON. As $ON\perp CD$, this becomes $ME\perp CD$. Hence, < (ME; CD) = 90º. Now, since the quadrilateral ABCD is cyclic, < DCA = < DBA, or, equivalently, < (CD; AC) = < EBA. Thus, < MEA = < (ME; AC) = < (ME; CD) + < (CD; AC) = 90º + < EBA = 90º - < ABE. Let $M_1$ be the circumcenter of triangle AEB. Then, this point $M_1$ lies on the perpendicular bisector of the side AB of this triangle. On the other hand, the point M, being the midpoint of the segment AB, also lies on this perpendicular bisector. Now, assume that $M_1\neq M$. Then, we thus conclude that the line $MM_1$ is the perpendicular bisector of the segment AB. On the other hand, since the point $M_1$ is the circumcenter of triangle AEB, the points A, E, B lie on a circle with center $M_1$; thus, by the central angle theorem for directed angles modulo 180º, we have $\measuredangle M_1EA=90^{\circ}-\measuredangle ABE$, so that $\measuredangle M_1EA=\measuredangle MEA$. Hence, the points E, M and $M_1$ are collinear, i. e. the line $MM_1$ passes through the point E. Since the line $MM_1$ is the perpendicular bisector of the segment AB, this yields that the perpendicular bisector of the segment AB passes through the point E, and thus AE = BE. This contradicts $AE\neq BE$. Hence, our assumption that $M_1\neq M$ was wrong, so we have $M_1=M$. Thus, the circumcenter of the triangle AEB is the midpoint of its side AB. This means that the triangle AEB is right-angled at E; in other words, $AC\perp BD$. This completes the proof of Assertion 2. Hence, Theorem 1 is proven. Thanks in advance to everyone who dares to read this all... darij

Can I just ask where did you find this problem??

In a national training set.. Bomb

darij grinberg wrote: What is more interesting (because much harder to prove) is that the converse also holds, i. e. we can prove: Theorem 1. Let ABCD be a (not necessarily convex) cyclic quadrilateral with circumcenter O, and let the diagonals AC and BD of the quadrilateral ABCD intersect at a point E. Then, the midpoints of the segments AB, CD and OE are collinear if and only if either AD = BC or $AC\perp BD$. darij A Nifty Solution Let $a,b,c,d$ be the complex numbers representing the vertices $A,B,C,D$ resp. of a quadrilateral inscribed in a unit circle. Let also $AC\cap BD=E$. Then the complex numbers $\frac{a+b}{2},\frac{c+d}{2},\frac{ac(b+d)-bd(a+c)}{2(ac-bd)}$ represents the midpoints of the segments $AB,CD,OE$ respectively. Also see, if we strictly impose that $AC,BD$ are diagonals and must intersect then the condition $AD=BC \iff AB\parallel CD$. Which in terms of complex numbers gives $\frac{a-b}{c-d}\in\mathbb{R}^*\iff \frac{a-b}{c-d}=\frac{\tfrac{1}{a}-\tfrac{1}{b}}{\tfrac{1}{c}-\tfrac{1}{d}}=\frac{a-b}{c-d}\cdot\frac{cd}{ab}\iff ab-cd=0$. Similarly we have $AC\perp BD \iff \frac{a-c}{b-d}\in i\mathbb{R}^*\iff \frac{a-c}{b-d}=-\frac{\tfrac{1}{a}-\tfrac{1}{c}}{\tfrac{1}{b}-\tfrac{1}{d}}=- \frac{a-c}{b-d}\cdot \frac{bd}{ac}\iff ac+bd=0$. Now the midpoints of $AB,CD,OE$ are collinear is equivalent to $\frac{\tfrac{ac(b+d)-bd(a+c)}{2(ac-bd)}-\tfrac{c+d}{2}}{\tfrac{c+d}{2}-\tfrac{a+b}{2}}\in\mathbb{R}^*$ which is equivalent to \begin{align*} &\frac{\tfrac{ac(b+d)-bd(a+c)}{2(ac-bd)}-\tfrac{c+d}{2}}{\tfrac{c+d}{2}-\tfrac{a+b}{2}}=\frac{\tfrac{\overline{ac(b+d)-bd(a+c)}}{2\left(\overline{ac-bd}\right)}-\tfrac{\overline{c+d}}{2}}{\tfrac{\overline{c+d}}{2}-\tfrac{\overline{a+b}}{2}}\\ \iff&\frac{ac(b-c)-bd(a-d)}{(ac-bd)(a+b-c-d)}=\frac{\tfrac{1}{ac}\cdot\tfrac{c-b}{bc}-\tfrac{1}{bd}\cdot\tfrac{d-a}{ad}}{\tfrac{bd-ac}{abcd}\cdot\left(\tfrac{1}{a}+\tfrac{1}{b}-\tfrac{1}{c}-\tfrac{1}{d}\right)}\\ \iff&\frac{ac(b-c)-bd(a-d)}{(a+b-c-d)}=\frac{ab(d^2(c-b)-c^2(d-a))}{(ad(b-c)+bc(a-d))}\\ \iff& (ac-bd)(ab-cd)(ac+bd)=0 \end{align*}Note throughout $ac-bd\neq 0$ as we strictly imposed that $AC,BD$ must intersect. Done!