Name the circle $k$ and denote $F$ the intersection of the tangentlines at $E$ and $F$ to the circle $k$. $F$ is the pole of $DE$ with respect to $k$, hence it lies on the polar $BC$ of $A\in DE$ with respect to $k$. As we have $XE=BD$ and therefore $\sphericalangle{XCE}=\sphericalangle{DCB}$ due to the parallel lines, proving that $XC$ is a median of triangle $\triangle{CDE}$ is equivalent to proving that $CB$, or $FC$, is a symmedian. We have by the sine law \[\left.\frac{\sin{\sphericalangle{DCB}}}{\sin{\sphericalangle{BCE}}}=\frac{\sin{\sphericalangle{FCD}}}{\sin{\sphericalangle{ECF}}}=\frac{\frac{\sin{\sphericalangle{CDF}}}{FC}\cdot FD}{\frac{\sin{\sphericalangle{FEC}}}{FC}\cdot FE}=\frac{\sin{\sphericalangle{CDF}}}{\sin{\sphericalangle{FEC}}}=\frac{\sin{\sphericalangle{CED}}}{\sin{\sphericalangle{EDC}}}\mbox{,}\qquad (1)\right.\] and if $CM$ with $M$ on $DE$ is a median of triangle $\triangle{CDE}$, we get by the sine law again that \[\left.\frac{\sin{\sphericalangle{MCE}}}{\sin{\sphericalangle{DCM}}}=\frac{\frac{\sin{\sphericalangle{MCE}}}{ME}}{\frac{\sin{\sphericalangle{DCM}}}{DM}}=\frac{\frac{\sin{\sphericalangle{CEM}}}{MC}}{\frac{\sin{\sphericalangle{MDC}}}{MC}}=\frac{\sin{\sphericalangle{CED}}}{\sin{\sphericalangle{EDC}}}\mbox{.}\qquad (2)\right.\] Combining (1) and (2) implies that $\sphericalangle{MCE}=\sphericalangle{DCB}$ and $\sphericalangle{DCM}=\sphericalangle{BCE}$, hence $FC$ really is the symmedian and we are done.

Dear Mathlinkers, another way is to consider "A chord CY parallel to DE" which would avoid the use of angles. Sincerely Jean-Louis

Let $O$ be the center of the circle. Note that $\angle AOC= \frac{1}{2} \angle BOC= \frac{1}{2} (\hat BD+ \hat DC)= \frac{1}{2} (\hat EX + \hat DC)= \angle DFC$. Thus, $AFOC$ is cyclic, so $\angle AFO=90$. So $OF \perp DE$, so $F$ is the midpoint of $DE$.

Dear Mathlinkers, my idea was to use the concept of symmedian...which would avoid the use of angles. Sincerely Jean-Louis

Note that BDEC is harmonic hence (D;E;C;B)=-1 Lets make a projection from X to AE: (D;E;C;B)=(D;E;XC∩AE;P∞)=-1 Hence XC bisects DE.

USAJMO 2011 P5 PAMO 2000 P5

Here is my solution for this problem Solution Let $M$ $\equiv$ $CX$ $\cap$ $DE$ We have: $\dfrac{MD}{ME}$ = $\dfrac{CD}{CE}$ . $\dfrac{XD}{XE}$ = $\dfrac{CD}{CE}$ . $\dfrac{BE}{BD}$ = 1 So: $M$ is midpoint of $DE$

Rename some points and use phantom points. We will solve the following equivalent problem: Quote: In $\triangle ABC$ with circumcircle $\Omega$, let $D$ be a point on $\Omega$ such that $AD$ is the $A$-symmedian. Let $T$ be the midpoint of $AD$ and let $BT$ meet the line through $C$ parallel to $AD$ at $E$. Show that $E$ lies on $\Omega$. We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Since $D$ is on the $A$-symmedian, it can be parameterized by $D=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (D)=0$. Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$, so plugging in $D$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$. So$$D=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$$Thus, the point at infinity along the line $A$-symmedian has coordinates $$P_\infty=(b^2+c^2:-b^2:-c^2).$$ Normalize $D$ by multiplying each component by $\frac{1}{-a^2+2b^2+2c^2}$. The midpoint formula on normalized $D$ and $A$ gives $T=(-a^2+b^2+c^2:b^2:c^2)$. Note that $E$ is the intersection of cevians $BT$ and $CP_\infty$. Points on $BT$ can be parameterized by $(-a^2+b^2+c^2:s:c^2)$ while points on $CP_\infty$ can be parameterized by $(b^2+c^2:-b^2:t)$. Hence, their intersection must have coordinates $$E=[(b^2+c^2)(-a^2+b^2+c^2):-b^2(-a^2+b^2+c^2):c^2(b^2+c^2)].$$It remains to check that $\text{Pow}_\Omega (E)=0$. Plugging $E$ into the circumcircle equation yeilds $$a^2[-b^2(-a^2+b^2+c^2)c^2(b^2+c^2)]+b^2[(b^2+c^2)(-a^2+b^2+c^2)c^2(b^2+c^2)]+c^2[(b^2+c^2)(-a^2+b^2+c^2)(-b^2)(-a^2+b^2+c^2)]$$which is clearly true after factoring.

This is a solution using complex numbers. Let $A=a, B=b, C=c, D=d, E=e$ and $X=x$. Using the circle as the unit circle: $a = \frac{2bc}{b+c}$ $\rightarrow$ $ $ $\overline{a} = \frac{\frac{2}{bc}}{\frac{1}{b}+\frac{1}{c}} = \frac{2}{b+c}$. $A, D, E$ are colineals: $\frac{a-d}{\overline{a}-\overline{d}} = \frac{d-e}{\overline{d}-\overline{e}}$ $\rightarrow$ $\frac{\frac{2bc}{b+c}-d}{\frac{2}{b+c}-\frac{1}{d}} = \frac{d-e}{\frac{1}{d}-\frac{1}{e}}$ $\rightarrow$ $\frac{d(2bc-bd-cd)}{2d-b-c}=-de$ $\rightarrow$ $2bc-bd-cd=-2de+be+ce$ $\rightarrow$ $\fbox{2bc+2de = (b+c)(d+e)}$ $\ldots$ $(\alpha)$ And $BX \parallel DE$: $\frac{b-x}{\overline{b}-\overline{x}}=\frac{d-e}{\overline{d}-\overline{e}}$ $\rightarrow$ $\frac{b-x}{\frac{1}{b}-\frac{1}{x}}=\frac{d-e}{\frac{1}{d}-\frac{1}{e}}$ $\rightarrow$ $-bx=-de$ $\rightarrow$ $x=\frac{de}{b} \ldots (i)$. Let $M$:middle point of $DE$ and $M=m$ $\rightarrow$ $m =\frac{d+e}{2}$ and $\overline{m}=\frac{\frac{1}{d}+\frac{1}{e}}{2}=\frac{d+e}{2de} \ldots (ii)$. We want to prove that $M \in XC$ $\leftrightarrow$ $x+c=m+xc\overline{m}$ Replacing with $(i)$ and $(ii)$: $\leftrightarrow$ $\frac{de}{b}+c=\frac{d+e}{2}+c(\frac{de}{b})\cdot(\frac{d+e}{2de})$ $\leftrightarrow$ $\frac{de+bc}{b}=\frac{d+e}{2}+\frac{cd+ce}{2b}$ $\leftrightarrow$ $\frac{de+bc}{b}=\frac{db+eb+cd+ce}{2b}$ $\leftrightarrow$ $2de+2bc=(d+e)(b+c)$ and this is true for $(\alpha)$. $\blacksquare$