<s>[INST] Geometry problem related to circle, triangles. Given acute triangle $\triangle ABC$ satisfying $|\overline{AB}| \ne |\overline{AC}|$. Let $D,E$, respectively, be the midpoints of $\overline{AB}, \overline {AC}$. Let $Q, P$ be the intersections of $(\triangle ADE)$ and $(\triangle BCD)$, $(\triangle ADE)$ and $(\triangle BCE)$, respectively. Prove that $|\overline{AP}| = |\overline{AQ}|$.
I have already tried using radical axis but still cannot figure out the solution. Please help me with this. Thanks.

 [/INST] We shall use Barycentric Coordinates with reference triangle $ABC$, that is $P=(x,y,z) \stackrel{def}{\iff} \vec P=x\vec A+y\vec B +z \vec C$ and $x+y+z=1$. Also, $(x:y:z)=(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z})$. Let $a=BC,\ b=CA,\ c=AB.$
The general equation of a circle(G.E.C.) is $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0.$
Let $T=PB\cap QC$. $AP=AQ\iff\angle AEP=\angle ADQ \iff \angle PBC= \angle QCB \iff K$ lies on perpendicular bisector of $BC$.
$A=(1:0:0),\ D=(1:1:0),\ E=(1:0:1)$. Plugging into G.E.C., we obtain $(ADE): -a^2yz-b^2zx-c^2xy+\frac{(x+y+z)}{2}(c^2y+b^2z)=0$. Also, $B=(0:1:0)$ and $C=(0:0:1)$. Therefore, $(BEC): -a^2yz-b^2zx-c^2xy+\frac{(x+y+z)}{2}(b^2x)=0$.
Without loss of generality, let $P=(1:y_1:z_1)$. Plugging $P$ into $(ADE)$ and $(BCE)$ and solving for $z_1$, $z_1=\frac{b^2-c^2}{2a^2}$. By symmetry, if $Q=(1:y_2:z_2)$, $y_2=\frac{c^2-b^2}{2a^2}$.
Therefore, $T=PB\cap QC=(1:\frac{c^2-b^2}{2a^2}:\frac{b^2-c^2}{2a^2})=(2a^2:c^2-b^2:b^2-c^2)$. Clearly, $T$ satisfies the equation of perpendicular bisector of $BC$ : 
 $a^2(y-z) +x(b^2-c^2)=0$
$\blacksquare$
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