Below is a "full" solution to the problem. A hint, if you don't feel quite ready to see the solution yet, is to consider the Miquel point of the complete quadrilateral $ABCD$.
Let points $A,B,C,D,E,F,M,P,Q,X,Y$ be defined as in the question. Define $\gamma$ to be the circumcircle of $ABCD$, and redefine $O$ to be its centre. Let the circle with centre in $Y$ through $X$ be called $\omega$.
Lemma 1 (Miquel Point of Cyclic Quadrilaterals): Let $Z$ be the intersection of lines $OX$ and $EF$. Then $Z$ is the miquel point of the complete quadrilateral $ABCD$. In particular, $Z$ is the image of $X$ under inversion with respect to $\gamma$.
Lemma 2 ($EFX$ is self-polar with respect to $\gamma$): $X$ and $Z$ lie on the normal line from $O$ to $EF$.
These two facts turn out to be sufficient in explaining the tangency in the question as follows:
Due to lemma 2 $\angle MZX=\angle MZO = \pi/2$. Therefore, since $MX$ is a diameter in $\omega$, $Z$ lies on $\omega$ by the converse of Thale's theorem.
But now lemma 1 tells us that $X$ and $Z$ are inverse images under inversion in $\gamma$, implying that $|OX||OZ|=r^2$, where $r$ is the radius of $\gamma$. Hence the power of $O$ with respect to $\omega$ is $r^2$. Now suppose a tangent to $\omega$ through $O$ intersect $\omega$ at $T$. Then by power of a point $|OT|^2=r^2$, so $T\in\gamma$. But $T\in\omega$ by assumption, so $T=P$ or $T=Q$.
Said in another way: the tangents from $O$ to $\omega$ are exactly the lines $OP$ and $OQ$.
Now you may realise that this is rather similar to what we want to prove, that is, the tangents from $Y$ to $\gamma$ are the lines $YP$ and $YQ$. It turns out that these two statements are actually equivalent. You may try to prove it yourself. The configuration is called orthogonal circles. At any rate, this explains the problem posed in the original post.
All of the concepts/lemmas that I have used are defined/proven thoroughly in chapters 8 to 10 of Evan Chen's Euclidean Geometry in Mathematical Olympiads . It is a good introduction to many of the more advanced techniques used in olympiad geometry, and I wholeheartedly recommend it if you are preparing for maths olympiads.
Any theorem worth using once is worth using four times!
Observe that $A, D, E, F, G$ lie on a circle $Γ$ centered at the midpoint $J$ of $AD$. Let $BD, CD$ meet $Γ$ again at $K, L$. Let the tangents to $Γ$ at $E, K$ meet at $M$, and the tangents at $F, L$ meet at $N$.
Since
$$\begin{multline*}\frac{∠KME}{2} = \frac{180° - ∠EJK}{2} = ∠KDE - 90° = ∠KBE \\ = ∠FCL = ∠FDL - 90° = \frac{180° - ∠LJF}{2} = \frac{∠FNL}{2},\end{multline*}$$
we see that
- $B, E, K$ lie on a circle $Γ_1$ centered at $M$;
- $C, F, L$ lie on a circle $Γ_2$ centered at $N$;
- there’s a diameter $\ell$ of $Γ$ that reflects $E, K, M, Γ_1$ to $F, L, N, Γ_2$.
Let $EL, FK, \ell$ meet at $I'$. Using Pascal’s theorem twice on $EEFKKL$ and $EFFKLL$, we find that $I'$ lies on $MN$, so it is in fact the midpoint of $MN$. Using Pascal’s theorem on $AELDKF$, we find that $I'$ lies on $BC$. This is enough to establish that $I' = I$, the midpoint of $BC$.
Finally, the result follows using Pascal’s theorem on $AEFKDG$.
Best Answer
There are two possibilities to construct the point $E$. Both start by constructing the point $X\in BD$, so that $X$ is the angle bisector of $\hat C$ in $\Delta ABC$.
Because of $\frac {XD}{XB}=\frac{CD}{CB}=\frac{DD}{EB}$ (angle bisector theorem, and the given equality of proportions) $ED$ is the angle bisector in $\Delta EDB$. So let us construct points $E$ with this property. Consider the circle $(X)$ centered in $X$ which is tangent to the line $AB$. Draw the two tangents from $D$ to this circle, they intersect $AB$ in two points, $E,E'$, both being possible choices for the point $E$ in the problem.
A second possibility uses the fact that the geometric locus of all points $P$ with given constant value for the ratio $k=PD/PB$ is a circle. This circle has the line $DB$ as symmetry. The simplest way to see this is to make use a system of coordinates with $D,B$ in $(-1,0)$ and $(1,0)$, then rewrite the given relation as $\displaystyle k^2=\frac{(x+1)^2+y^2}{(x-1)^2+y^2}$. In our case, this circle passes through $X, X'$ the points where the inner, respectively outer angle bisector of $\hat C$ in $\Delta ABC$ intersect $BD$. This is a symmetry axis, so $XX'$ is a diameter in this circle, let $\Xi$ be its center, the mid point of $XX'$. We denote this circle by $(\Xi)$. It intersects $AB$ in two points $E,E'$. (We let $E$ be the point closer to $B$.) See also
https://en.wikipedia.org/wiki/Circle#Circle_of_Apollonius
The second possibility is maybe better to work with, so we restate the problem equivalently, avoiding the annoying metric condition. As i always proceed in such situations, my solution is not the quick solution going straightforward to the needed conclusion, instead, all "interesting" properties in the given geometric constellation (related or not to the target property) are listed and shown. A long time i walked well with this strategy, the understanding of the problems is optimal, alternative solutions are possible. So let us state and prove the following...
Problem: Let $\Delta ABC$ be a triangle. We introduce the following points.
Then we have:
A picture:
Proofs:
(1) Because $XX'$ is a diameter in $(\Xi)$ we have $XE'\perp X'E'$, $XE\perp X'E$, so $XE'$, $X'E$ are heights in $\Delta TXX'$. Which is the third height? $EX$ bisects $\widehat{DEB}$ and $EXX'E'$ cyclic. This implies: $$ \widehat{DEX}= \widehat{XEB}= \widehat{BX'E'}= \widehat{XX'T} \ , $$ so $EDX'X$ cyclic, so $\widehat{TDX'}=\widehat{TEX'}=90^\circ$. This was (1).
But in the same place we insert a picture of the Euler circle in the triangle $\Delta TXX'$, a circle passing through the feet of the heights, $E, E', D$, but also throught the mid point $\Xi$ of the base $XX'$.
(4) This is the usual property of the height, $$ \widehat{TDE}= \widehat{HDE}= \widehat{HXE}= \widehat{E'XT}= 90^\circ-\hat{T}= \widehat{EX'T}= \widehat{HX'E'}= \widehat{HDE'}= \widehat{TDE'} $$
Alternatively consider the mid point of $TH$, and use the fact that $\Xi$ and this point determine a diameter in the Euler circle, which is perpendicular on $EE'$. As an observation, we may write using the Euler circle and the circle $(\Xi)$: $$ \widehat{EDE'}= \widehat{E\Xi E'}= 2\widehat{EXE'}= 2\widehat{EX'E'}= 2\widehat{ECE'}= \ . $$
(3) It is more or less "the same situation" as in (1), but the points $F,F'$ are more complicated. Since $D$ is the midpoint of $AC$, we have $DO\perp ADC$. The diameter of $(O)$ on the line $DO$ is $LL'$. So $\Delta LL'F$ and $\Delta LL'F'$ have a right angle opposed to $LL'$. Then $LF'$, $L'F$ are two heights in $\Delta ULL'$. Which is the third height? We want to show it is $UD$. To have an immediate parallel to (1) here is the triangle $\Delta ULL'$ and its Euler circle, showing what we want to show:
An analytic solution is given, since i also need the ingredients for (2).
We use barycentric coordinates in the triangle $\Delta ABC$. Its side lengts are $a,b,c$ say, by usual notation. The computations will use standard notations, please consult
Barycentric coordinates for the impatient, Max Schindler and Evan Chen
for the details.
Some coordinates and equations are immediate. $$ \begin{aligned} A &= [1:0:0]=(1,0,0)\ ,\\ B &= [0:1:0]=(0,1,0)\ ,\\ C &= [0:0:1]=(0,0,1)\ ,\\ D &= \frac 12(A+C)=[1:0:1]=\frac 12(1,0,1)\ ,\\ X &= \frac {BC}{BC+CD}D+\frac {CD}{BC+CD}B =\frac{2a}{2a+b}\cdot\frac 12(1,0,1)+\frac{b}{2a+b}\cdot(0,1,0)\\ &=[a:b:a]=\frac 1{2a+b}(a,b,a)\ ,\\ &\text{... or use formally $(D-X)=-\frac b{2a}(B-X)$, i.e. $\frac{XD}{XB}=-\frac b{2a}$.}\\ &\text{... and for $X'$ formally $(D-X)=+\frac b{2a}(B-X)$, i.e. $\frac{XD}{XB}=+\frac b{2a}$,}\\ &\text{... so $X'$ is formally obtained from $X$ via $b\to-b$,}\\ X' &= \frac {BC}{BC+CD}D+\frac {CD}{BC+CD}B =\frac{2a}{2a+b}\cdot\frac 12(1,0,1)+\frac{b}{2a+b}\cdot(0,1,0)\\ &=[a:b:a]=\frac 1{2a+b}(a,b,a)\ ,\\ (\Xi) &=\text{the circle through $C,X,X'$, it has the equation}\\ 0 &= -a^2yz -b^2zx -c^2xy+(ux+vy)(x+y+z)\ ,\qquad\text{ where}\\ u &=\frac 1{4a^2-b^2}\cdot b^2(a^2-c^2)\ ,\\ v &=\frac 1{4a^2-b^2}\cdot a^2(2a^2-b^2+2c^2)\ ,\\ E,E'&\text{ are the two points $E_\pm$ in the intersection of $(\Xi)$ with $AB$ $(z=0)$,}\\ E_+ &= [1:m_+^E:0]=(x_+^E, y_+^E, 0)=(x_+^E, m_+^Ex_+^E, 0) =\frac 1{1+m_+^E}(1,m_+^E,0) \ ,\\ E_- &= [1:m_-^E:0]=(x_-^E, y_-^E, 0) =(x_-^E, m_-^Ex_-^E, 0) =\frac 1{1+m_-^E}(1,m_-^E,0) \ ,\\ &\qquad\text{where $m_\pm$ are the two roots of the equation in $M$}\\ &\qquad\text{obtained by setting $z=0$ and $y=Mx$ in $(\Xi)$:}\\ 0 &= -c^2 m+(u+mv)(1+m) \\ &=vm^2-(c^2-u-v)m+u\ ,\text{ so}\\ \Sigma &=m_+ + m_- = \frac 1v{(c^2-u-v)}=-\frac{2(a^2-c^2)}{2a^2-b^2+2c^2}\ ,\\ \Pi &=m_+ m_- = \frac uv=\frac{b^2(a^2-c^2)}{a^2(2a^2-b^2+2c^2)} =\frac{b^2}{2a^2}\Sigma\ ,\\ L,L' &= L_\pm =\text{ intersection of $(O)$ with perpendicular in $D$ on $AB$,}\\ L_+ &=[a(a+c) : -b^2: c(a+c)]\ ,\\ L_- &=[a(a-c) : -b^2: c(c-a)]\ ,\\ &=\qquad\text{ (and observe that $+c\leftrightarrow -c$ exchanges formally $L_+\leftrightarrow L_-$.)}\\ S &=\text{ intersection of $AB$ ($z=0$) with perpendicular in $D$ on $DB$}\\ &=\text{ solution of $x+y+z=1$, $z=0$, and}\\ &=\text{ (EFFT) for displacements $\frac 12(2x-1,2y,2z-1)$ and $\frac 12(1, -2, 1)$, so}\\ S &= [2a^2-b^2+2c^2:c^2-a^2:0]\ . \end{aligned} $$ The discriminant of the equation in $M$ is not a square in the fraction field of the ring $\Bbb[a,b,c]$, so we try not to write $m$ explicitly. Now we compute $F_\pm$, which is the solution of the equations in $(x,y,z)$: $$ \left\{ \begin{aligned} 0 &=a^2yz+b^2zx+c^2xy\ ,\\ m_\pm =\frac{y^E_\pm}{x^E_\pm} & =\frac yx\ ,\\ 1 &=x+y+z\ . \end{aligned} \right. $$ The solution of this system, where the second equation is $y=mx$ is $$ [ma^2+b^2\ :\ m(ma^2+b^2)\ :\ -mc^2]\ . $$ We obtain $F_\pm$ by setting $m_\pm$ instead of $m$. The equations for the lines $LF_+$ and $L'F_-$ are: $$ LF_+\ :\ \begin{vmatrix} a(a-c) & -b^2 & c(c-a)\\ m_+a^2+b^2 & m_+(m_+a^2+b^2) & -m_+c^2\\ x & y & z \end{vmatrix} =0 \ ,\qquad L'F_-\ :\ \begin{vmatrix} a(a+c) & -b^2 & c(c+a)\\ m_-a^2+b^2 & m_-(m_-a^2+b^2) & -m_-c^2\\ x & y & z \end{vmatrix} =0 $$ And we want to show they intersect in a point $U\in AC$, so $y_U=0$. We add the equation $x+y+z=1$ to the above two, determining $U$, and the second component is by Kramer's rule: $$ \begin{vmatrix} \color{blue}{ \begin{vmatrix} -b^2 & c(c-a)\\ m_+(m_+a^2+b^2) & -m_+c^2\\ \end{vmatrix}} & 0 & \color{magenta}{ \begin{vmatrix} a(a-c) & -b^2 \\ m_+a^2+b^2 & m_+(m_+a^2+b^2) \\ \end{vmatrix}} \\ \color{magenta}{ \begin{vmatrix} -b^2 & c(c+a)\\ m_-(m_-a^2+b^2) & -m_-c^2\\ \end{vmatrix}} & 0 & \color{blue}{ \begin{vmatrix} a(a+c) & -b^2 \\ m_-a^2+b^2 & m_-(m_-a^2+b^2) \\ \end{vmatrix}} \\ 1 & 1 & 1 \end{vmatrix} $$ (divided by the "determinant of the system"). So it is enough to show that the product
$ \color{blue}{P_\searrow}$ of the blue terms is the product $ \color{magenta}{P_\nearrow}$ of the purple terms. (So the products are invariant w.r.t. the "Galois substitution" $m_+\leftrightarrow m_-$ done in the same time with $c\leftrightarrow -c$.) $$ \begin{aligned} \color{blue}{P_\searrow} &= \color{blue}{ \begin{vmatrix} -b^2 & c(c-a)\\ m_+(m_+a^2+b^2) & -m_+c^2\\ \end{vmatrix} \cdot \begin{vmatrix} a(a+c) & -b^2 \\ m_-a^2+b^2 & m_-(m_-a^2+b^2) \\ \end{vmatrix}} \\ &= m_+\cdot \begin{vmatrix} -b^2 & c(c-a)\\ m_+a^2+b^2 & -c^2\\ \end{vmatrix} \cdot (m_-a^2+b^2) \cdot \begin{vmatrix} a(a+c) & -b^2 \\ 1 & m_- \\ \end{vmatrix} \\ &= m_+\cdot \begin{vmatrix} -b^2 & c(c-a)\\ m_+a^2 & -ac\\ \end{vmatrix} \cdot (m_-a^2+b^2) \cdot \begin{vmatrix} a(a+c) & -b^2 \\ 1 & m_- \\ \end{vmatrix} \\ &= m_+\cdot ac\begin{vmatrix} a(a-c) & -b^2\\ 1 & m_+\\ \end{vmatrix} \cdot (m_-a^2+b^2) \cdot \begin{vmatrix} a(a+c) & -b^2 \\ 1 & m_- \\ \end{vmatrix} \\ &= ac\;\color{red}{m_+\;(m_-a^2+b^2)} \;(a(a-c)m_+ + b^2)\;(a(a+c)m_- + b^2)\ ,\\[3mm] &\qquad\text{ and similarly}\\[3mm] \color{magenta}{P_\nearrow} &= \color{magenta}{ \begin{vmatrix} -b^2 & c(c+a)\\ m_-(m_-a^2+b^2) & -m_-c^2\\ \end{vmatrix} \cdot \begin{vmatrix} a(a-c) & -b^2 \\ m_+a^2+b^2 & m_+(m_+a^2+b^2) \\ \end{vmatrix}} \\ &= m_- \begin{vmatrix} -b^2 & c(c+a)\\ m_-a^2+b^2 & -c^2\\ \end{vmatrix} \cdot (m_+a^2+b^2) \begin{vmatrix} a(a-c) & -b^2 \\ 1 & m_+ \\ \end{vmatrix} \\ &= m_- \begin{vmatrix} -b^2 & c(c+a)\\ m_-a^2 & ac\\ \end{vmatrix} \cdot (m_+a^2+b^2) \begin{vmatrix} a(a-c) & -b^2 \\ 1 & m_+ \\ \end{vmatrix} \\ &= m_-\cdot a(-c) \begin{vmatrix} a(c+a) &-b^2 \\ 1 & m_- \\ \end{vmatrix} \cdot (m_+a^2+b^2) \begin{vmatrix} a(a-c) & -b^2 \\ 1 & m_+ \\ \end{vmatrix} \\ &=ac\; \color{red}{(-m_-)\;(m_+a^2+b^2)}\;(a(a+c)m_- + b^2)\;(a(a-c)m_+ + b^2) \ . \end{aligned} $$ We have the equaltiy of the terms marked in red, because of $$ m_+\;(m_-a^2+b^2) + m_-\;(m_+a^2+b^2) =2a^2\underbrace{m_+m_-}_\Pi +b^2\underbrace{(m_+ + m_-)}_\Sigma=0\ , $$ since $\Pi$ is $-\frac{b^2}{2a^2}\Sigma$. The other two factors are matching each other mot-a-mot.
$\square$
To recapitulate, we showed that $UD$ is the height of $\Delta ULL'$ (and the other two heights are $FL'$ and $F'L$), by computing the intersection $LF_+\cap L'F_-$, and showing that its $y$-component vanishes. (We can implant then either $F,F'$ or $F',F$ for the values $F_+,F_-$.) This shows (3) analytically.
(5) This is parallel to (4), it is the usual property of the height, $$ \widehat{ADF}= \widehat{RDF}= \widehat{RLF}= \widehat{F'LF}= 90^\circ-\hat{U}= \widehat{F'L'F}= \widehat{F'L'R}= \widehat{F'DR}= \widehat{F'DA} $$ or we can use the Euler circle in the same way as above, considering its diameter from $O$ to the mid point of $UR$, here $R$ is the orthocenter in $\Delta ULL'$. As an observation, we may write using the Euler circle and the circumcircle $(O)$: $$ \widehat{FDF'}= \widehat{FOF'}= 2\widehat{FLF'}= 2\widehat{FL'F'}= 2\widehat{FCF'}\ . $$
(6) This is what the OP is asking for. So far we have used intentionally two different marks for the pairs of equal (unsigned) angles in $D$: $$ \widehat{EDT}= \widehat{E'DT} \text{ and } \widehat{FDA}= \widehat{F'DA}\ . $$ They are equal, because using the circles $(O)$ and $(\Xi)$ they "united in $C$, being equal to $$ \gamma:= \widehat{ECE'}= \widehat{F'CF'}\ . $$
A rotation around $D$ with angle $\gamma$ brings the rays $DE$ into $DT$, and $DF$ into $DA$, so $\widehat{EDF}=\widehat{TDA}$. A further $\gamma$-rotation around $D$ shows
$\widehat{EDF}=\widehat{TDA}=\widehat{E'DF'}$. Using $DB\perp DT$ and $DL\perp DA$, a $90^\circ$ rotation around $D$ gives $\widehat{BDL}=\widehat{TDA}$.
This shows the OP: $$ 90^\circ-\widehat{CDB}= \widehat{BDL}= \widehat{EDF}= \widehat{E'DF'}\ . $$
It remains (2), a bonus in the given constellation. We use again barycentric coordinates. Recall that the points $F_\pm$ are $$ [m_\pm a^2+b^2\ :\ m_\pm(m_\pm a^2+b^2)\ :\ -m_\pm c^2]\ . $$ Then the line $F_+F_-$ has the equation $$ \begin{vmatrix} m_+ a^2+b^2\ &\ m_+(m_+a^2+b^2)\ &\ -m_+c^2\\ m_- a^2+b^2\ &\ m_-(m_-a^2+b^2)\ &\ -m_-c^2\\ x & y & z \end{vmatrix} =0\ . $$ Let $S=(x_S,y_S,z_S)$ be the intersection $F_+F_-\cap AB$. From $S\in AB$ we have $z_S=0$, so $x_S+y_S=1$, and expanding the above determinant w.r.t. the third line: $$ \begin{vmatrix} m_+(m_+a^2+b^2)\ &\ m_+\\ m_-(m_-a^2+b^2)\ &\ m_- \end{vmatrix} x_S - \begin{vmatrix} m_+ a^2+b^2\ &\ m_+\\ m_- a^2+b^2\ &\ m_- \end{vmatrix} y_S =0\ . $$ (The last column was simplified, it appears without the factor $-c^2$.) In the coefficient of $x_S$ we linearly extract on the lines the factors $m_\pm$. Then the new second column $1,1$ is used to linearly get rid of $+b^2,+b^2$ from the first column. Then in the new first column we extract the factor $a^2$.
In the coefficient of $y_S$ we use the second column $m_+,m_-$ to linearly get rid of $m_+a^2,m_-a^2$ from the first column. Then in the new first column we extract the factor $a^2$. This gives: $$ m_+m_-a^2 \begin{vmatrix} m_+\ &\ 1\\ m_-\ &\ 1 \end{vmatrix} x_S - b^2 \begin{vmatrix} 1\ &\ m_+\\ 1\ &\ m_- \end{vmatrix} y_S =0\ . $$ Recall that $m_+m_-=\Pi=\frac uv=\frac{b^2(a^2-c^2)}{a^2(2a^2-b^2+2c^2)} $, so after simplifying the above to $m_+m_-a^2x_S+b^2y_S=0$ and plugging in the formula for $m_+m_-=\Pi$ we obtain: $$ (a^2-c^2)x_S+(2a^2-b^2+2c^2)y_S=0\ , \ x_S+y_S=0\ , $$ and the solution is $$ y_S=-\frac{a^2-c^2}{a^2-b^2+3c^2}\ ,\ x_S=1-y_S\ , $$ so we compute the displacement vectors $$ \begin{aligned} DS &= S-D = (x_S,y_S,0)-\frac 12(1,0,1) =\left(x_S-\frac 12,y_S,-\frac 12\right) \\ &=\left(\frac 12-y_S,y_S,-\frac 12\right) \sim[1-2y_S\ :\ 2y_S\ :\ -1]\ ,\\ DB &= B-D =(0,1,0)-\frac 12(1,0,1) \\ &=-\frac 12(1,-2,1) \sim[1\ :\ -2\ :\ 1]\ . \end{aligned} $$ The perpendicularity $DS\perp DB$ is then equivalent to (EFFT): $$ a^2(2y_S\cdot 1+(-1)\cdot(-2) ) + b^2((1-2y_S)\cdot 1+(-1)\cdot 1 ) + c^2((1-2y_S)\cdot (-2)+2y_S\cdot1 ) =0\ . $$ which is $$ y_S(2a^2-2b^2+6c^2)+(2a^2-2c^2)=0\ . $$ Which is true.
$\square$
A final remark: All points (1) to (6) are now proven. Using barycentric coordinates was leading to "straightforward" solutions for (2), (3). (Using a computer algebra system, "straightforward" becomes straightforward. Here, there is some effort to type the solution, on paper things are simpler.)
If analytic = computational solutions "must be" avoided, then one needs a proof for (3), or at least for a simple part in it, for instance it is enough to show $FF'DO$ cyclic. (Note that $O$ is on the side bisector of $FF'$.) For the bonus point (2) projective geometry is maybe the path, but i could not find a proof (using Desargues, Pappus, Pascal, et caetera) in time.
But also note that barycentric coordinates are a strong tool in competitions (Olympiads, although then one should also prove formulas like EFFT for a full point harvest). Here is then an explicit example how this works in practice. No details were omitted.
I will still search for synthetic solutions, but now i have to submit, time.