Prove intersection between side length and tangent to circumcircle at opposite vertex is collinear with points on perpendicular bisectors of sides Let $ABC$ be a triangle with $AB\neq BC$. Point $E$ lies on the perpendicular bisector of $AB$ such that $BE\perp BC$. Point $F$ lies on the perpendicular bisector of $AC$ such that $CF\perp BC$. Let $D$ be the point of intersection between $BC$ and the tangent to the circumcircle of $\triangle ABC$ at $A$. Prove $D,E,F$ are collinear.
So there is a pretty easy solution with complex numbers, here is a sketch of that (I haven't worked out all the details and I may have made a mistake somewhere, but there is definitely a way to use complex numbers without too much computation):
Let $\odot(ABC)$ be the unit circle. Then we may write $D=AA\cap BC=\frac{a^2(b+c)-2abc}{a^2-bc}$. From WLOG $a=1$, we can get $D=\frac{b+c-2bc}{1-bc}$. But from the perpendicularity conditions, we get $\frac{e}{1-b},\frac{f}{1-c}\in i\mathbb{R}$. From here we deduce $\frac{e}{\overline{e}}=b,\frac{f}{\overline{f}}=c$ Plugging in to $D$, we can get $d=\frac{e\overline{f}+\overline{e}f-2ef}{\overline{e}\overline{f}-ef}$. From here it remains to check 
$ \left| \begin{array}{ccc}
e & \overline{e} & 1 \\
f & \overline{f} & 1 \\
d & \overline{d} & 1 \end{array} \right| = 0. $ 
I am looking for a solution using synthetic geometry (e.g. no coordinates, vectors, etc.) I tried to find one but didn't get very far, so I'd be happy if anyone could help me out. All I see is we can define $D'=BC\cap EF$, and then we have a homothety centered at $D'$ mapping $F$ to $E$ and $C$ to $B$, and it remains to show that $D'=D$. Any observations that may lead to a solution are appreciated. Thanks!
 A: For the sake of avoiding configuration issues, we will use directed angles.
Let $G$ be the second point on $(ABC)$ such that $DG$ is tangent to $(ABC)$, and let $H$ be the second point of intersection of the circles centered at $E$ and $F$ that both pass through $A$. Clearly $E$ and $F$ now lie on the perpendicular bisector of $AH$, so we will prove that $D$ does too by proving $|DA| = |DH|$, immediately proving $D$, $E$ and $F$ are collinear.

However, note that $|DG| = |DA|$ by equal tangent chords, so it is sufficient to show that $H$ and $G$ are reflections in $BC$ such that $|DG| = |DH|$ and hence $|DA| = |DH|$.
Now note that $ABGC$ is a harmonic quadrilateral, so equivalently $G$ is the second point on $(ABC)$ such that $GB/GC = AB/AC$ (this can be proven directly using the $aa$-similarities $\triangle DBG \sim \triangle DGC$ and $\triangle DBA \sim \triangle DAC$ and the fact that $|DG| = |DA|$). As $H$ lies on the same side of $BC$ as $A$ (as it is the intersection of two circles that are fully on the same side of $BC$ as $A$ except tangency points $B$ and $C$), it now suffices to prove that $$HB/HC = AB/AC,$$ as reflecting $H$ in $BC$ would then give for the reflection $H'$ that $H'B/H'C = AB/AC$ and furthermore $H'$ would be on $(ABC)$ as then $\angle BH'C = \angle CHB = \angle HCB + \angle CBH = \angle HAC + \angle BAH = \angle BAC$, using directed angles and the fact that $(BAH)$ and $(CAH)$ are both tangent to $BC$.
For this home-stretch, we let $\alpha = \angle BAH = \angle CBH$ and $\beta = HAC = \angle HCB$. Now note that $AH$ is an $A$-median. Indeed, letting $M$ be the intersection of $BC$ and $AH$ we note that $M$ lies on the radical axis of $(BAH)$ and $(CAH)$ so $MB^2 = MH \cdot MA = MC^2$, such that $M$ is indeed the midpoint of $M$. This implies that areas $[BAM] = [MAC]$, so the sine area formula yields
$$
  1 = \frac{[BAM]}{[MAC]} = \frac{AB \cdot AM \cdot \sin \alpha}{AM \cdot AC \cdot \sin \beta},
$$
such that
$$
  AB/AC = \sin \beta / \sin \alpha.
$$
However, using the law of sines within $\triangle BHC$ also yields
$$
  HB/HC = \sin \beta/\sin \alpha.
$$
Hence $HB/HC = AB/AC$, so $G$ and $H$ are reflections in $BC$, so $|DH| = |DG| = |DA|$, so $D$ lies on the perpendicular bisector of $AH$, so $D$, $E$ and $F$ are collinear.
