Let $\ell \cap A O=\{K\}$ and $G$ be the other end point of the diameter of $\Gamma$ through $A$. Then $D, C, G$ are collinear. Moreover, $E$ is the orthocenter of triangle $A D G$. Therefore $G E \perp A D$ and $G, E, B$ are collinear.
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As $\angle C D F=\angle G D K=\angle G A C=\angle G F C, F G$ is tangent to the circumcircle of triangle $C F D$ at $F$. As $\angle F B E=\angle F B G=\angle F A G=\angle G F K=\angle G F E, F G$ is also tangent to the circumcircle of $B F E$ at $F$. Hence the circumcircles of the triangles $C F D$ and $B F E$ are tangent at $F$.
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1. Given Setup and Definitions:
- Let $A$, $B$, and $C$ be points on a circle $\Gamma$ with center $O$.
- $\angle ABC > 90^\circ$.
- $D$ is the intersection of line $AB$ with the line perpendicular to $AC$ at $C$.
- Line $l$ through $D$ is perpendicular to $AO$.
- $E$ is the intersection of $l$ with line $AC$.
- $F$ is the intersection of $\Gamma$ with $l$ between $D$ and $E$.

2. Intersection Points and Concyclic Points:
- Let $\overline{DE}$ meet $\Gamma$ again at $L$.
- Let $\overline{AO}$ meet $\Gamma$ again at $R$.
- Note that $R$ lies on line $\overline{CD}$.
- Let $\overline{DE}$ meet $\overline{AO}$ at $K$.
- Points $E$, $C$, $R$, and $K$ are concyclic.

3. Power of a Point:
- By the Power of a Point theorem, we have:
$$\overline{DA} \cdot \overline{DB} = \overline{DC} \cdot \overline{DR} = \overline{DE} \cdot \overline{DK}$$
- This implies that $A$, $B$, $E$, and $K$ are concyclic.

4. Angle Calculation:
- Since $A$, $B$, $E$, and $K$ are concyclic, we have:
$$\angle ABE = 90^\circ$$

5. Symmetry and Tangency:
- Points $F$ and $L$ are symmetric with respect to line $\overline{RA}$.
- Therefore, we have:
$$\angle BFD = \overarc{BL} = \overarc{BR} + \overarc{RL} = \overarc{BR} + \overarc{FR}$$
- This can be rewritten using angles:
$$\angle BFD = \angle BAK + \angle FLR = \angle BED + \angle FCD$$
- This proves that the circumcircles of $\triangle BFE$ and $\triangle CFD$ are tangent at point $F$.