Find the largest integer value of x Problem:
In the below figure, $E$ is the excenter of $\triangle ABC$ such that $\angle ACB=5x.$ Also, point $F$ is on $AC$ such that $\angle CBF=x$ and point $D$ is on the extension of $BC$ such that $DE\perp AE,\; DE\perp BH,\;$ where $H$ is the intersection of $DE$ and $BF$. Find the largest integer value of $x.$

My progress:

I marked all the angles I found but they all result in the same equation
$2\alpha =180^{\circ} - 5x \implies \boxed{\alpha = 90^{\circ} - \frac{5x}{2}}$
 A: Observe,
$$\angle BHD=\angle AED=90^{\circ}\implies AE\parallel BH$$
$$\;\angle EAF=\angle BFA=\angle FBC+\angle FCB=6x$$
Extend ray $BA$ to some point $P$. Since $E$ is the ex-center and $AE\parallel BH$,
$$\angle FBA=\angle EAP=\angle EAC=6x$$
Therefore, in $\triangle ABC$, $\angle ABC=7x\;$ and $\;\angle ACB=5x$.
$$\angle ABC+\angle ACB<180^{\circ}\implies 12x<180^{\circ}\implies x<15^{\circ} $$
The largest possible integer value of $x$ is $14^{\circ}$. In order to show that that $x=14^{\circ}$ is possible, chase angles and note that all given constraints are satisfied.
A: Let $\alpha,\beta,\gamma$ by so that the angles in $A,B,C$ of the given triangle are $2\alpha$, $2\beta$, $2\gamma$. All the time we will use
$$\alpha+\beta+\gamma=90^\circ\ .$$
Then we have in the picture the following order of introducing points, and then calculating angles joined with these points in the same order. (This is the reason for asking this in the comments.)


*

*We start with $\Delta ABC$.

*We construct (ignoring the alphabet) first the point $E$, the $B$-excenter, usually denoted by $I_B$.

*We draw the perpendicular line in $E$ on $AE$, and consider its intersection with $BC$, denoted by $D$.

*Let also $H$ be the projection of $B$ on $ED$.

*The problem also tacitly gives the information / the relation on the two angles marked with $x$ and $5x$. (The last angle is five times the first one.) Somewhere it should be noted that the angles are measured in degrees.

Now the question is which is the maximal value of the angle $x$ (measured in degrees), knowing it is an integer.

Solution:
We compute as in the picture first the angles in $\Delta ACE$, starting with those in $A,C$. Then for the angle in $E$ there remains $\alpha+\gamma$, so the complement is $\widehat{CED}=\beta$. In $\Delta CED$ we know two angles, those in $C,E$ (introduced as points before $D$), so the angle in $D$ is $\alpha+2\gamma$. So $x$ is the complement of this angle in $D$, which is $x=\beta-\gamma>0$. (This condition is soon satisfied.)
The given relation for the two angles $x$ and $5x$ becomes now:
$$
5(\beta-\gamma)=5x=2\gamma\ ,
$$
in other words $5\beta=7\gamma$. The angles of the triangle are thus so that
$$
90^\circ=\alpha+\beta+\frac 57\beta>\frac{12}7\beta\ .
$$
So
$$
x=\beta-\gamma=\beta-\frac 57\beta=\frac 27\beta
=\frac 16\cdot
\frac{12}7\beta
<\frac 16\cdot 90^\circ=15^\circ\ .
$$
So the maximal value in degrees which is an integer is $\color{blue}{14^\circ}$.
$\square$

The above solution is constructive, shows how to take $\alpha,\beta,\gamma$ explicitly to realize the maximum.
