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The first difficulty is to draw the figure: how to draw a point tangent to a line and passing through two given points? A common strategy in such cases is to reverse the drawing: start by placing points $A, B$, and $D$, then deduce the tangent $m$ and the point $C$.
Let $\Omega$ and $\Omega'$ be the circumcircles of $ABD$ and $DEF$, and $O$ and $O'$ their centers. Our hard-earned figure suggests that $\Omega'$, $m$, and $(AC)$ are concurrent. This is not surprising: indeed, it is to be shown that $\mathcal{C}'$ is the circle with diameter $[FD]$, or that the midpoint of $[AC]$, which we will denote as $A'$, belongs to $\Omega'$.
Now, the three points $A'$, $E$, and $F$ are the midpoints of the segments $[CA]$, $[CB]$, and $[CK]$, respectively. The circumcircle of these three points, which we wish to prove is $\Omega'$, is therefore the image of $\Omega$ under the homothety centered at $C$ with a ratio of $1/2$. Given the figure, it is therefore necessary to prove, if we denote $K'$ as the symmetric point of $K$ with respect to $O$, that this homothety maps $K'$ to $D$, i.e., that $D$ is the midpoint of $[K'C]$. By construction, we already know that $\widehat{KDK'} = 90^\circ$.
We will now prove that $(KD)$ is the bisector of $\widehat{CKK'}$, using angle chasing. First, since $AKO$ is isosceles at $O$ and $(AK)$ and $(DO)$ are perpendicular to $m$, we know that

$$(KC, KK') = (KA, KO) = (AO, AK) = (OA, OD)$$

Furthermore, since $m$ is tangent to $\Omega$, we also know that

$$(KC, KD) = (KA, KD) = (DA, DA') = (DA, CA) + 90^\circ = (DA, DO) + 90^\circ.$$

Since $ADO$ is isosceles at $O$, we conclude as desired that

$$2(KC, KD) = 2(DA, DO) = (DA, DO) + (AO, AD) = (OA, OD) = (KC, KK').$$

Comment from the examiners: The problem was adequately addressed, with several copies reaching far and even concluding with simple observations about the figure, such as the right angle $\widehat{KDC}$. Few students thought to introduce the point $K'$ from the official solution; although introducing this point was not strictly necessary, all students who did so subsequently solved the problem and received excellent grades.
It can therefore be judicious to introduce new points, but many do so without a specific reason, simply to express an angle in a new way. Here, there are very strong reasons to want to introduce the point $K'$, and it is necessary to convince oneself of the possible utility of a point before introducing it, at the risk of getting lost in unnecessary considerations.