The proof given below uses Wallace-Simson theorem and several properties of the Wallace-Simson line of a point with respect to a triangle. This theorem asserts that the feet of the perpendiculars dropped from an arbitrary point on the circumcircle of a triangle to its sides are collinear. The line that passes through the mentioned feet is called the Wallace-Simson line. Three of the many properties of this line are,
$\quad\space\space\text{i.$\enspace$The Wallace-Simson line of a vertex of the triangle is the altitude of the triangle dropped}\\\qquad\enspace \text{ from that vertex.}$
$\quad\space\text{ii.$\enspace$If $P$ and $Q$ are points on the circumcircle of the triangle, then the angle between the}\\\qquad\space \text{ Wallace-Simson lines of $P$ and $Q$ is half the angle of the arc $PQ$.}$.
$\quad\text{iii.$\enspace$If $P$ is a point on the circumcircle of the triangle, its Wallace-Simson line passes through}\\\qquad\space \text{ the midpoint of the line segment joining $P$ and the orthocentre of the triangle.}$
As shown in the $\mathrm{Fig.\space 1}$, line $\pmb{l}$ is an arbitrary line which passes through the orthocentre $H$ of an arbitrary triangle $ABC$ It cuts the sides $BC$, $CA$, and $AB$ of the $\triangle ABC$ at $A_1$, $B_1$, and $C_1$ respectively.
For the time being, assume that we used properties (i) and (ii) to geometrically determine the location of point $K$, so that its Wallace-Simson line is parallel to the line $\pmb{l}$, which intersects the sides $BC$, $CA$, and $AB$ of the $\triangle ABC$ at $U$, $V$, and $W$ respectively. The corresponding geometrical construction is given at the end of this text.
According to the property (iii), point $M$ is the midpoint of $HK$. We complete the scenario by joining points $A_1$, $B_1$, $C_1$, $U$, $V$, and $W$ to $K$. The Wallace-Simson line meets $A_1K$, $B_1K$, and $C_1K$ at $D$, $E$, and $F$, respectively. Now, our aim is to show that $A_1K$, $B_1K$, and $C_1K$ are the reflections of line $\pmb{l}$ on the sides $BC$, $CA$, and $AB$ of the $\triangle ABC$ respectively. For brevity, we let $\measuredangle WKC_1 =\phi$.
Consider the $\triangle C_1KH$. By construction, $HC_1$ is parallel to $MF$. Therefore, since $M$ is the midpoint of $HK$, it follows from the converse of midpoint theorem, that $C_1F=FK$, i.e., $F$ is the midpoint of $C_1K$.
According to the Wallace-Simson theorem, $KW$ is perpendicular to $AB$ and that makes the $\triangle C_1WK$ a right-angled triangle.
$${\large{\pmb{\therefore}}}\quad\space\measuredangle KC_1W =90^o-\phi.$$
Since $F$ is the midpoint of the hypotenuse of this triangle, we have, $FW=C_1F=FK$, which makes $\triangle WFC_1$ an isosceles triangle.
$${\large{\pmb{\therefore}}}\quad\space \measuredangle C_1WF = \measuredangle FC_1W = 90^o-\phi.$$
Since line $\pmb{l}$ is parallel to the Wallace-Simson line, we shall write,
$$\measuredangle WC_1G=\measuredangle C_1WF=90^o-\phi.$$
Since, $\measuredangle WC_1G=\measuredangle C_1WF$, $AB$ is the angle bisector of $\measuredangle KC_1G$.
Applying similar reasoning to appropriate geometrical objects in the diagram, one can prove $A_1K$ and $B_1K$ are the reflections of line $\pmb{l}$ on the sides $BC$ and $CA$ respectively
$\underline{\text{The Construction }}:$
We describe here concisely how to locate the point on the circumcircle of a given triangle $ABC$, the Wallace-Simson line of which is parallel to a
given line $FG$.
We drop the perpendicular $AD$ from the vertex $A$ of the $\triangle ABC$ to its opposite side $BC$. We do this because of the property (i). In other words, we know that $A$ is on the circumcircle of the $\triangle ABC$ and its Wallace-Simson line is the constructed altitude $AD$. Therefore, $A$ serves as the base of our construction. Assume that $AD$ meets the given line $FG$ at $E$ while the two lines making an angle $\theta$ as shown in $\mathrm{Fig.\space 2}$.
After drawing the circumcircle of $\triangle ABC$, we join its center $O$ to $A$. According to the property (ii), we need to find a point $P$ on the circumcircle so that the chord $AP$ must subtend an angle equal to $2\theta$ at the center $O$. This can be achieved by constructing the radius $OP$ so that $\measuredangle POA=2\theta$. After drawing perpendiculars from $P$ to any two sides of the $\triangle ABC$, join their feet to obtain the sought Wallace-Simson line $ST$ of point $P$.
Best Answer
Hint :
Tangent segments from an outside point to a circle are equal in length. So we have $AB=AF$, $CB=CD$ and $ED=EF$. Now the proof should be straightforward.