Dear Mathlinkers, the problem seems to me not clear. But, I can be wrong... All the circles pass through A and D.??? Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, the problem seems to me not clear. But, I can be wrong... All the circles pass through A and D.??? Sincerely Jean-Louis Dear Jean-Louis The problem says that $D$ is not a constant point, it varies, we have to find a constant point different from $A.$

hint

I have a solution for this problem. First we will construct $S$ be constant point such that $S\in \Gamma_D $. Let $H\in AB$ such that $CA=CH$ and let $w$ be the circle pass through $A,H$ and tangent to $CA,CH$. $w\cap (B,H,C)=S(S\neq H)$. Now we will prove that $ADSF$ is cyclic. We have $\angle BHC=180^0-\angle AHC=180^0-\angle BAC=\angle BDC$ so $BHDC$ is cyclic but $BHSC$ is cyclic hence $BHDS$ is cyclic. Thus $\angle SDF=180^0-\angle BDS=180^0-\angle BHS=\angle SHA=\angle SAF$ so $ADSF$ is cyclic and the problem is done

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$B \equiv AF \cap ED$ and $ C \equiv AE \cap FD$ are conjugate points WRT $\Gamma_D.$ Then it follows that circle $ (M)$ with diameter $ BC$ is orthogonal to $\Gamma_D.$ Thus, inversion through pole $A$ with power equal to the power of $A$ WRT $(M)$ takes $ (M)$ into itself and $\Gamma_D$ into a straight line $\gamma$ orthogonal to $(M),$ due to the conformity $\Longrightarrow$ $ M \in \gamma.$ Hence, $\Gamma_D$ passes through the image of $M$ under the referred inversion, i.e. the orthogonal projection of the orthocenter of $\triangle ABC$ on $ AM.$

Luis González wrote: $B \equiv AF \cap ED$ and $ C \equiv AE \cap FD$ are conjugate points WRT $\Gamma_D$ What do you mean by 'conjugate points with respect to a circle' and why does it imply orthogonality?

shoki wrote:

hint

Shoki ,could you explain better ?

Let circle (AFE) intersect (BDC) at D and G and line AG intersect (BDC) at X easy to see that XB || AE and XC || AF , let AX intersect BC at point M easy to see that M is midpoint of BC and (BDC) is constant , so G is fixed

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Quote: Let $H$ be the orthocenter.Let $H_b,H_c$ be the feet of perpendiculars from $B,C$ to $AC,AB$. the problem is equivalent to prove that $\frac{H_bE}{H_cF}$ is constant. @cnyd, see here

Hi ; This Point Is Also On $AM$ Where $M$ Is The Midpoint Of $BC$ Such That If We Connect It To $I$ Then We Have $IK$ Is Perpendicular To $AM$ Best Regard

We use barycentric coordinates. Let $BC = a, CA = b, AB = c$, and let $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$. Let $D = (p:q:r)$. We claim that $\Gamma_D$ always passes through $P(a^2:b^2+c^2-a^2:b^2+c^2-a^2)$, which is independent of $D$. Since $E$ and $F$ are the traces of $D$ onto $CA$ and $AB$, we have $E = (p:0:r)$ and $F = (p:q:0)$. We are given that $A, E, D, F$ are concyclic; that is, the equations of the circumcircles of $AED$ and $AFD$ are equal. \begin{align*}\odot AED: a^2yz+b^2zx+c^2xy - (x+y+z)(\frac{a^2r(p+r) - b^2rp + c^2p(p+r)}{(p+q+r)(p+r)}y + \frac{b^2p}{p+r}z) = 0\end{align*}\begin{align*}\odot AFD: a^2yz+b^2zx+c^2xy - (x+y+z)(\frac{c^2p}{p+q}y+\frac{a^2q(p+q) + b^2p(p+q) - c^2pq}{(p+q+r)(p+q)}z) = 0\end{align*} Therefore $(p+q)(a^2r(p+r)-b^2rp+c^2p(p+r)) = (p+q+r)(p+r)c^2p$, or $c^2p(p+r) = (p+q)(a^2(p+r)-b^2p)$. Similarly, $b^2p(p+q+r)(p+q) = (p+r)(a^2q(p+q)+b^2p(p+q)-c^2pq)$, or $b^2p(p+q)= (p+r)(a^2(p+q) - c^2p)$. Multiplying the two yields \[a^2(p+r)(p+q) = b^2p(p+q)+c^2p(p+r)\] Solving for $c^2p(p+r)$ and plugging this into the equation of the circumcircle of $AED$ yields \[a^2yz+b^2zx+c^2xy - (x+y+z)(a^2y + \frac{b^2p}{p+r}(z-y))) = 0\] Substituting $P$ into this equation yields \begin{align*}a^2(b^2+c^2-a^2)(b^2+c^2-a^2)+b^2(b^2+c^2-a^2)(a^2)+c^2(a^2)(b^2+c^2-a^2) - (-a^2+2b^2+2c^2)(a^2(b^2+c^2-a^2)) = 0,\end{align*}which is true. Therefore the circumcircle of $AEDF$ always passes through $P$, as claimed. This was a lot more manageable than I'd thought. I didn't have to use Mathematica to expand the last expression, since the assumed condition $y = z$ allowed me to find the $x$ and $y$-coordinates of $P$ in the first place.

Luis González wrote: $B \equiv AF \cap ED$ and $ C \equiv AE \cap FD$ are conjugate points WRT $\Gamma_D.$ Then it follows that circle $ (M)$ with diameter $ BC$ is orthogonal to $\Gamma_D.$ Alternatively, we can finish here by noting that $\Gamma_D$ passes through the second limit point of the coaxial system, different from $A.$ But of course a characterization of this point gives a nicer solution.

let $ABHC$ be a parallelogram and let $k=\odot HBC$ meet $AH$ again at a fixed point $L$. $\angle BDC=\angle EDF=180-\angle BAC=180-\angle BHC$ so $D\in k$. Now $\angle DLA=180-\angle DLH=\angle HBD=\angle AEB$ so $L\in \Gamma_{D}$ Q.E.D

We do not use barycentric coordinates. Clearly the locus of $D$ is tha arc of a circle $\omega$ with chord $BC$ such that $\angle BDC=\pi -\angle A$. Fix any point $D$ on this locus, and let $\Gamma_D$ intersect $\omega$ again at $X$. I claim $X$ is the desired common point. Take another point $D'$ on the locus, and let the circumcircle of $\triangle AD'X$ intersect $AB$ and $AC$ at $F'$ and $E'$ respectively. If we show $C, D',$ and $F'$ are collinear then similarly $B, D',$, and $E'$ are collinear, so $\Gamma_D'$ goes through $X$. We have $\angle F'D'X+\angle XD'C=\angle F'AX+\pi-\angle XDC=\pi$, so $C, D',$ and $F'$ are collinear as desired, and we are done. QED. (of course, leader's solution is much more elegant and gives a nice characterization of $X$, but it is harder to find) This was a lot more manageable than I'd thought. I didn't have to use Mathematica to expand the last expression, since I did not use barycentric coordinates in the first place.